KC Sinha Mathematics Solution Class 10 Chapter 4 Trigonometric Ratios and Identities Exercise 4.1

Q1 Q2 Q3 Q4 A Q4 B Q4 C Q5 A Q5 B Q6 Q7 A Q7 B Q8 Q9 Q10 A Q10 B Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 A Q18 B Q19 Q20 Q21 Q22 A Q22 B Q23 Q24 A Q24 B Q24 C Q25 Q26 A Q26 B Q27 Q28 Q29 Q30 Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40 Q41 Q42 A Q42 B Q42 C Q42 D Q42 E Q42 F Q43 A Q43 B Q43 C Q43 D Q44 Q45 Q46 Q47 Q48 Q49 Q50 Q51 Q52 Q53 Q54 Q55 Q56

Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4

Exercise 4.1

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Question 1

From the given figure, find the value of the following:

(i) sin C
(ii) sin A
(iii) cos C
(iv) cos A
(v) tan C
(vi) tan A
Sol :
(i) Sin C
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
So, here θ = C
Side opposite to ∠C = AB = 3
Hypotenuse = AC = 5
So, $\sin C=\frac{A B}{A C}=\frac{3}{5}$

(ii) Sin A
So, here θ = A
The side opposite to ∠A = BC = 4
Hypotenuse = AC = 5
So, $\sin A=\frac{B C}{A C}=\frac{4}{5}$

(iii) Cos C
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
So, here θ = C
Side adjacent to ∠C = BC = 4
Hypotenuse = AC = 5
So, $\cos C=\frac{B C}{A C}=\frac{4}{5}$

(iv) Cos A
Here, θ = A
Side adjacent to ∠A = AB = 3
Hypotenuse = AC = 5
So, $\cos A=\frac{A B}{A C}=\frac{3}{5}$

(v) tan C
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
So, here θ = C
Side opposite to ∠C = AB = 3
Side adjacent to ∠C = BC = 4
So, $\tan C=\frac{A B}{B C}=\frac{3}{4}$

(vi) tan A
here θ = A
Side opposite to ∠A = BC = 4
Side adjacent to ∠A = AB = 3
So, $\tan A=\frac{A B}{B C}=\frac{4}{3}$

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Question 2 

From the given figure, find the value of :

(i) tan θ
(ii) cos θ

Sol :
(i) tan θ
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Side opposite to θ = AB = 4
Side adjacent to θ = BC = 3
So, $\tan \theta=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{4}{3}$

(ii) cos θ
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to θ = BC = 3
Hypotenuse = AC = 5
So, $\cos \theta=\frac{B C}{A C}=\frac{3}{5}$

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Question 3 

From the given figure, find the value of

(i) sin θ
(ii) tan θ
(iii) tan A – cot C

 

Sol :
(i) sin θ
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to θ = BC = ?
Hypotenuse = AC = 13
Firstly we have to find the value of BC.
So, we can find the value of BC with the help of Pythagoras theorem.
According to Pythagoras theorem,
(Hypotenuse)² = (Base)² + (Perpendicular)²
⇒ (AB)² + (BC)² = (AC)²
⇒ (12)² + (BC)² = (13)²
⇒ 144 + (BC)² = 169
⇒ (BC)² = 169–144
⇒ (BC)² = 25
⇒ BC =√25
⇒ BC =±5
But side BC can’t be negative. So, BC = 5
Now, BC = 5 and AC = 13
So, $\sin \theta=\frac{B C}{A C}=\frac{5}{13}$

(ii) tan θ
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Side opposite to θ = BC = 5
Side adjacent to θ = AB = 12
So, $\tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$

(iii) tan A – cot C
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
and
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
tan A
Here, θ = A
Side opposite to ∠A = BC = 5
Side adjacent to ∠A = AB = 12
So, $\tan \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$
Cot C
Here, θ = C
Side adjacent to ∠C = BC = 5
Side opposite to ∠C = AB = 12
So, $\cot C=\frac{B C}{A B}=\frac{5}{12}$
So, $\tan A-\cot C=\frac{5}{12}-\frac{5}{12}=0$

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Question 4 A 

In ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine

a. sin A, cos A
b. sin C, cos C
Sol :

(i)
(a) sin A
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
So, here θ = A
Side opposite to ∠A = BC = 7
Hypotenuse = AC = ?
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem.

According to Pythagoras theorem,
(Hypotenuse)² = (Base)² + (Perpendicular)²
⇒ (AB)² + (BC)² = (AC)²
⇒ (24)² + (7)² = (AC)²
⇒ 576 + 49 = (AC)²
⇒ (AC)² = 625
⇒ AC =√625
⇒ AC =±25
But side AC can’t be negative. So, AC = 25cm
Now, BC = 7 and AC = 25
So, $\sin A=\frac{B C}{A C}=\frac{7}{25}$
Cos A
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
So, here θ = A
Side adjacent to ∠A = AB = 24
Hypotenuse = AC = 25
So,$\cos A=\frac{A B}{A C}=\frac{24}{25}$
(b) sin C

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
So, here θ = C
The side opposite to ∠C = AB = 24
Hypotenuse = AC = 25
So, $\sin C=\frac{A B}{A C}=\frac{24}{25}$
Cos C
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
So, here θ = C
Side adjacent to ∠C = BC = 7
Hypotenuse = AC = 25
So, $\cos C=\frac{B C}{A C}=\frac{7}{25}$

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Question 4 B 

Consider ∆ACB, right angled at C, in which AB = 29 units, BC = 21 units and ∠ABC=θ. Determine the values of

a. cos² θ+ sin² θ
b. cos² θ – sin² θ
Sol :

(a) Cos²θ +sin² θ
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem.
According to Pythagoras theorem,
(Hypotenuse)² = (Base)² + (Perpendicular)²
⇒ (AC)² + (BC)² = (AB)²
⇒ (AC)² + (21)² = (29)²
⇒ (AC)² = (29)² – (21)²
Using the identity a² –b² = (a+b) (a – b)
⇒ (AC)² = (29–21)(29+21)
⇒ (AC)² = (8)(50)
⇒ (AC)² = 400
⇒ AC =√400
⇒ AC =±20
But side AC can’t be negative. So, AC = 20units
Now, we will find the sin θ and cos θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
In ∆ACB, Side opposite to angle θ = AC = 20
and Hypotenuse = AB = 29
So, $\sin \theta=\frac{A C}{A B}=\frac{20}{29}$
Now, We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$

In ∆ACB, Side adjacent to angle θ = BC = 21
and Hypotenuse = AB = 29
So, $\cos \theta=\frac{B C}{A B}=\frac{21}{29}$
So$\cos ^{2} \theta+\sin ^{2} \theta=\left(\frac{21}{29}\right)^{2}+\left(\frac{20}{29}\right)^{2}$
$=\frac{441+400}{29 \times 29}$
$=\frac{841}{841}$
=1
Cos²θ +sin² θ = 1
(b) Cos²θ – sin² θ
Putting values, we get
$\cos ^{2} \theta-\sin ^{2} \theta=\left(\frac{21}{29}\right)^{2}-\left(\frac{20}{29}\right)^{2}$
$=\frac{441-400}{29 \times 29}$
$=\frac{41}{841}$

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Question 4 C 

In ∆ABC, ∠A is a right angle, then find the values of sin B, cos C and tan B in each of the following :

a. AB = 12, AC = 5, BC = 13
b. AB = 20, AC = 21, BC = 29
c. BC = √2, AB = AC = 1
Sol :
Given that ∠A is a right angle.

(a) AB = 12, AC = 5, BC = 13
To Find : sin B, cos C and tan B
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Here, θ = B
Side opposite to angle B = AC = 5
Hypotenuse = BC =13
So, $\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{5}{13}$
Now, Cos C
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = C
Side adjacent to angle C = AC = 5
Hypotenuse = BC =13
So, $\cos C=\frac{A C}{B C}=\frac{5}{13}$
Now, tan B
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Here, θ = B
The side opposite to angle B = AC = 5
The side adjacent to angle B = AB = 12
So, $\tan \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{5}{12}$
(b) AB = 20, AC = 21, BC = 29
To Find: sin B, cos C and tan B
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Here, θ = B
The side opposite to angle B = AC =21
Hypotenuse = BC =29
So, $\sin B=\frac{A C}{B C}=\frac{21}{29}$
Now, Cos C
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = C
Side adjacent to angle C = AC = 21
Hypotenuse = BC = 29
So,$\cos C=\frac{A C}{B C}=\frac{21}{29}$

Now, tan B
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Here, θ = B
The side opposite to angle B = AC = 21
The side adjacent to angle B = AB = 20
So, $\tan \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{21}{20}$

(c) BC =√2, AB = AC = 1
To Find: sin B, cos C and tan B
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Here, θ = B
The side opposite to angle B = AC =1
Hypotenuse = BC =√2
So$\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{1}{\sqrt{2}}$
Now, Cos C
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = C
Side adjacent to angle C = AC = 1
Hypotenuse = BC = √2
So, $\cos C=\frac{A C}{B C}=\frac{1}{\sqrt{2}}$
Now, tan B
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Here, θ = B
The side opposite to angle B = AC = 1
The side adjacent to angle B = AB = 1
So,$\tan B=\frac{A C}{A B}=\frac{1}{1}=1$

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Question 5 A 

Find the value of the following : (a) sin θ (b) cos θ (c) tan θ from the figures given below :

Sol :
Firstly, we give the name to the midpoint of BC i.e. M
BC = BM + MC = 2BM or 2MC
⇒ BM = 5 and MC = 5
Now, we have to find the value of AM, and we can find out with the help of Pythagoras theorem.
So, In ∆AMB
⇒ (AM)² + (BM)² = (AB)²
⇒ (AM)² + (5)² = (13)²
⇒ (AM)² = (13)² – (5)²
Using the identity a² –b² = (a+b) (a – b)
⇒ (AM)² = (13–5)(13+5)
⇒ (AM)² = (8)(18)
⇒ (AM)² = 144
⇒ AM =√144
⇒ AM =±12
But side AM can’t be negative. So, AM = 12
a. sin θ

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
In ∆AMB
Side opposite to θ = AM = 12
Hypotenuse = AB=13
So, $\sin \theta=\frac{A M}{A B}=\frac{12}{13}$
So, $\sin \theta=\frac{12}{13}$
b. cos θ
We know that, $\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
In ∆AMB
The side adjacent to θ = BM = 5
Hypotenuse = AB = 13
So, $\cos \theta=\frac{B M}{A B}=\frac{5}{13}$
So, $\cos \theta=\frac{5}{13}$
c. tan θ

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
In ∆AMB
Side opposite to θ = AM = 12
The side adjacent to θ = BM = 5
So, $\tan \theta=\frac{\mathrm{AM}}{\mathrm{BM}}=\frac{12}{5}$
So, $\tan \theta=\frac{12}{5}$

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Question 5 B 

Find the value of the following : (a) sin θ (b) cos θ (c) tan θ from the figures given below :

Sol :
Firstly, we have to find the value of XM and we can find out with the help of Pythagoras theorem
So, In ∆XMZ
⇒ (XM)² + (MZ)² = (XZ)²
⇒ (XM)² + (16)² = (20)²
⇒ (XM)² = (20)² – (16)²
Using the identity a² –b² = (a+b) (a – b)
⇒ (XM)² = (20–16)(20+16)
⇒ (XM)² = (4)(36)
⇒ (XM)² = 144
⇒ XM =√144
⇒ XM =±12
But side XM can’t be negative. So, XM = 12
Now, In ∆XMY we have the value of XM and MY but we don’t have the value of XY.
So, again we apply the Pythagoras theorem in ∆XMY
⇒ (XM)² + (MY)² = (XY)²
⇒ (12)² + (5)² = (XY)²
⇒ 144 + 25 = (XY)²
⇒ (XY)² = 169
⇒ XY =√169
⇒ XY =±13
But side XY can’t be negative. So, XY = 13
a. sin θ
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
In ∆XMY
Side opposite to θ = MY = 5
Hypotenuse = XY = 13
So, $\sin \theta=\frac{\mathrm{MY}}{\mathrm{XY}}=\frac{5}{13}$
b. cos θ
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
In ∆XMY
Side adjacent to θ = XM = 12
Hypotenuse = XY = 13
So,$\cos \theta=\frac{\mathrm{XM}}{\mathrm{XY}}=\frac{12}{13}$
c. tan θ
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
In ∆XMY
The side opposite to θ = MY = 5
Side adjacent to θ = XM = 12
So, $\tan \theta=\frac{\mathrm{MY}}{\mathrm{XM}}=\frac{5}{12}$

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Question 6 

In ∆PQR, ∠Q is a right angle PQ = 3, QR = 4. If ∠P=α and ∠R=β, then find the values of

(i) sin α (ii) cos α
(iii) tan α (iv) sin β
(v) cos β (vi) tan β
Sol :

Given : PQ = 3, QR = 4
⇒ (PQ)² + (QR)² = (PR)²
⇒ (3)² + (4)² = (PR)²
⇒ 9 + 16 = (PR)²
⇒ (PR)² = 25
⇒ PR =√25
⇒ PR =±5
But side PR can’t be negative. So, PR = 5
(i) sin α
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Here, θ = α
The side opposite to angle α = QR =4
Hypotenuse = PR =5
So, $\sin \alpha=\frac{4}{5}$
(ii) cos α
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = α
The side adjacent to angle α = PQ =3
Hypotenuse = PR =5
So, $\cos \alpha=\frac{3}{5}$
(iii) tan α
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Here, θ = α
Side opposite to angle α = QR =4
Side adjacent to angle α = PQ =3
So,$\tan \alpha=\frac{4}{3}$
(iv) sin β
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Here, θ = β
The side opposite to angle β = PQ =3
Hypotenuse = PR =5
So, $\sin \beta=\frac{3}{5}$
(v) cos β
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = β
Side adjacent to angle β = QR =4
Hypotenuse = PR =5
So, $\cos \beta=\frac{4}{5}$
(vi) tan β
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Here, θ = β
Side opposite to angle β = PQ =3
Side adjacent to angle β = QR =4
So, $\tan \beta=\frac{3}{4}$

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Question 7 A 

If $\sin \theta=\frac{4}{5}$ ,then find the values of cos θ and tan θ.

Sol :
Given: $\sin \theta=\frac{4}{5}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{4}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{4}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4}{5}$
Let,
Perpendicular =AB =4k
and Hypotenuse =AC =5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
In right angled ∆ ABC, we have
⇒ (AB)² + (BC)² = (AC)²
⇒ (4k)² + (BC)² = (5k)²
⇒ 16k² + (BC)² = 25k²
⇒ (BC)² = 25 k² –16 k²
⇒ (BC)² = 9 k²
⇒ BC =√9 k²
⇒ BC =±3k
But side BC can’t be negative. So, BC = 3k
Now, we have to find the value of cos θ and tan θ
We know that,
$\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
The side adjacent to angle θ or base = BC =3k
Hypotenuse = AC =5k
So, $\cos \theta=\frac{3 \mathrm{k}}{5 \mathrm{k}}=\frac{3}{5}$
Now,
We know that,$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Perpendicular = AB =4k
Base = BC =3k
So, $\tan \theta=\frac{4 k}{3 k}=\frac{4}{3}$

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Question 7 B 

If $\sin \mathrm{A}=\frac{3}{4}$ ,calculate cos A and tan A.

Sol :
Given: Sin A $=\frac{3}{4}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{4} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{4}$
Let,
Side opposite to angle θ = BC =3k
and Hypotenuse = AC =4k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (AB)² + (3k)² = (4k)²
⇒ (AB)² + 9k² = 16k²
⇒ (AB)² = 16 k² – 9 k²
⇒ (AB)² = 7 k²
⇒ AB =k√7
So, AB = k√7
Now, we have to find the value of cos A and tan A
We know that,
$\cos \theta=\frac{\text { Side adjacent to angle } \theta}{\text { hypotenuse }}$
Here, θ = A
The side adjacent to angle A = AB =k√7
Hypotenuse = AC =4k
So,$\cos A=\frac{k \sqrt{7}}{4 k}=\frac{\sqrt{7}}{4}$
Now,
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
The side opposite to angle A = BC =3k
The side adjacent to angle A = AB =k√7
So,$\tan \mathrm{A}=\frac{3 \mathrm{k}}{\mathrm{k} \sqrt{7}}=\frac{3}{\sqrt{7}}$

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Question 8 

If $\sin \theta=\frac{3}{5}$ , then find the values cos θ and tan θ.

 

Sol :
Given:$\sin \theta=\frac{3}{5}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{3}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{5}$
Let,
Perpendicular =AB =3k
and Hypotenuse =AC =5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (BC)² = (5k)²
⇒ 9k² + (BC)² = 25k²
⇒ (BC)² = 25 k² – 9 k²
⇒ (BC)² = 16 k²
⇒ BC =√16 k²
⇒ BC =±4k
But side BC can’t be negative. So, BC = 4k
Now, we have to find the value of cos θ and tan θ
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = BC =4k
Hypotenuse = AC =5k
So, $\cos \theta=\frac{4 \mathrm{k}}{5 \mathrm{k}}=\frac{4}{5}$
Now, tan θ
We know that,
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Perpendicular = AB =3k
Base = BC =4k
So, $\tan \theta=\frac{3 \mathrm{k}}{4 \mathrm{k}}=\frac{3}{4}$

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Question 9 

If $\cos \theta=\frac{4}{5}$ , then find the value of tan θ.

Sol :

We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Or $\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}$
$\cos \theta=\frac{4}{5} \Rightarrow \frac{B}{H}=\frac{4}{5} \Rightarrow \frac{B C}{A C}=\frac{4}{5}$
Let,
Base =BC = 4k
Hypotenuse =AC = 5k
Where, k ia any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (AB)² + (4k)² = (5k)²
⇒ (AB)² + 16k² = 25k²
⇒ (AB)² = 25 k² –16 k²
⇒ (AB)² = 9 k²
⇒ AB =√9 k²
⇒ AB =±3k
But side AB can’t be negative. So, AB = 3k
Now, we have to find tan θ
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Side opposite to angle θ = BC =4k
Side adjacent to angle θ = AB =3k
So,$\tan \theta=\frac{4 k}{3 k}=\frac{4}{3}$

---

Question 10 A 

If $\tan \theta=\frac{3}{4}$ then find the values of cos θ and sin θ.

 

Sol :

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \theta=\frac{3}{4}$

$\Rightarrow \frac{P}{B}=\frac{3}{4}$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{3}{4}$
Let,
The side opposite to angle θ =AB = 3k
The side adjacent to angle θ =BC = 4k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (4k)² = (AC)²
⇒ (AC)² = 9 k²+16 k²
⇒ (AC)² = 25 k²
⇒ AC =√25 k²
⇒ AC =±5k
But side AC can’t be negative. So, AC = 5k
Now, we will find the sin θ and cos θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AB = 3k
and Hypotenuse = AC = 5k
So, $\sin \theta=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$
Now, We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = BC = 4k
and Hypotenuse = AC = 5k
So,$\cos \theta=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

---

 Question 10 B 

If tan A= 4/3. Find the other trigonometric ratios of the angle A.

Sol :

We know that,

$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Here, θ = A

$\tan \mathrm{A}=\frac{4}{3}$

$\Rightarrow \frac{P}{B}=\frac{4}{3}$

$\Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{4}{3}$

Let,
The side opposite to angle A =BC = 4k
The side adjacent to angle A =AB = 3k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (4k)² = (AC)²
⇒ (AC)² = 9 k² +16 k²
⇒ (AC)² = 25 k²
⇒ AC =√25 k²
⇒ AC =±5k
But side AC can’t be negative. So, AC = 5k
Now, we will find the sin A and cos A
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle A = BC = 4k
and Hypotenuse = AC = 5k
So, $\sin A=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

Now, We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle A = AB = 3k
and Hypotenuse = AC = 5k

So, $\cos A=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$
Now, we find other trigonometric ratios
$\operatorname{cosec} \mathrm{A}=\frac{1}{\sin \mathrm{A}}$
$=\frac{1}{\frac{4}{5}}$
$=\frac{5}{4}$

$\sec \mathrm{A}=\frac{1}{\cos \mathrm{A}}$
$=\frac{1}{\frac{3}{5}}$
$=\frac{5}{3}$

$\cot A=\frac{1}{\tan A}$
$=\frac{1}{\frac{4}{3}}$
$=\frac{3}{4}$

---

Question 11 

If cot $\theta=\frac{12}{5}$, then find the value of sin θ.

Sol :

We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{12}{5} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{12}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{12}{5}$
Let,
Side adjacent to angle θ =AB = 12k
The side opposite to angle θ =BC = 5k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (12k)² + (5k)² = (AC)²
⇒ (AC)² = 144 k² +25 k²
⇒ (AC)² = 169 k²
⇒ AC =√169 k²
⇒ AC =±13k
But side AC can’t be negative. So, AC = 13k
Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 5k
and Hypotenuse = AC = 13k
So, $\sin \theta=\frac{B C}{A C}=\frac{5 k}{13 k}=\frac{5}{13}$

---

Question 12 

If tan $\theta=\frac{5}{12}$, then find the value of cos θ.

Sol :

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \theta=\frac{5}{12} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{5}{12} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$
Let,
The side opposite to angle θ =BC = 5k
The side adjacent to angle θ =AB = 12k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (12k)² + (5k)² = (AC)²
⇒ (AC)² = 144 k² +25 k²
⇒ (AC)² = 169 k²
⇒ AC =√169 k²
⇒ AC =±13k
But side AC can’t be negative. So, AC = 13k
Now, We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = AB = 12k
and Hypotenuse = AC = 13k
So, $\cos \theta=\frac{A B}{A C}=\frac{12 k}{13 k}=\frac{12}{13}$

---

Question 13 

If sin $\theta=\frac{12}{13}$, then find the value of cos θ and tan θ.

 

Sol :
Given: $\sin \theta=\frac{12}{13}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{12}{13} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{12}{13} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12}{13}$
Let,
Side opposite to angle θ = 12k
and Hypotenuse = 13k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (12k)² + (BCk)² = (13)²
⇒ 144 k² + (BC)² = 169 k²
⇒ (BC)² = 169 k² –144 k²
⇒ (BC)² = 25 k²
⇒ BC =√25 k²
⇒ BC =±5k
But side BC can’t be negative. So, BC = 5k
Now, we have to find the value of cos θ and tan θ
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = BC =5k
Hypotenuse = AC =13k
So, $\cos \theta=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$
Now, tan θ
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
side opposite to angle θ = AB =12k
Side adjacent to angle θ = BC =5k
So, $\tan \theta=\frac{12 \mathrm{k}}{5 \mathrm{k}}=\frac{12}{5}$

---

Question 14 

If tan θ =0.75, then find the value of sin θ.

 

Sol :

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$

Given: tan θ =0.75
$\Rightarrow \tan \theta=\frac{75}{100}=\frac{3}{4}$
$\tan \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{3}{4} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3}{4}$
Let,
The side opposite to angle θ =BC = 3k
The side adjacent to angle θ =AB = 4k
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (4k)² + (3k)² = (AC)²
⇒ (AC)² = 16 k² +9 k²
⇒ (AC)² = 25 k²
⇒ AC =√25 k²
⇒ AC =±5k
But side AC can’t be negative. So, AC = 5k
Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 3k
and Hypotenuse = AC = 5k
So, $\sin \theta=\frac{B C}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$

---

Question 15 

If tan B= √3, then find the values of sin B and cos B.

 

Sol :

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Given: tan B = √3
$\Rightarrow \tan \mathrm{B}=\frac{\sqrt{3}}{1}$
$\tan \mathrm{B}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$
Let,
Side opposite to angle B =AC = √3k
The side adjacent to angle B =AB = 1k
where k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (1k)² + (√3k)² = (BC)²
⇒ (BC)² = 1 k² +3 k²
⇒ (BC)² = 4 k²
⇒ BC =√2 k²
⇒ BC =±2k
But side BC can’t be negative. So, BC = 2k
Now, we will find the sin B and cos B
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle B = AC = k√3
and Hypotenuse = BC = 2k
So, $\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$
Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle B = AB =1k
Hypotenuse = BC =2k
So, $\cos \mathrm{B}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1 \mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}$

---

Question 16 

If $\tan \theta=\frac{\mathrm{m}}{\mathrm{n}}$, then find the values of cos θ and sin θ.

 

Sol :

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Here, $\tan \theta=\frac{\mathrm{m}}{\mathrm{n}}$
So, Side opposite to angle θ =AC = m
The side adjacent to angle θ =AB = n
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (n)² + (m)² = (BC)²
⇒ (BC)² = m² + n²
⇒ BC =√ m² + n²
So, BC =√(m² + n²)
Now, we will find the sin B and cos B
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AC = m
and Hypotenuse = BC =√(m² + n²)
So, $\sin \theta=\frac{A C}{B C}=\frac{m}{\sqrt{m^{2}+n^{2}}}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = AB =n
Hypotenuse = BC =√(m² + n²)
So, $\cos \theta=\frac{A B}{B C}=\frac{n}{\sqrt{m^{2}+n^{2}}}$

---

Question 17 

If sin θ = √3 cos θ, then find the values of cos θ and sin θ.

 

Sol :
Given : sin θ =√3cos θ
$\Rightarrow \frac{\sin \theta}{\cos \theta}=\sqrt{3}$
⇒ tan θ =√3

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
and tan θ = √3
$\Rightarrow \tan \theta=\frac{\sqrt{3}}{1}$
$\tan \theta=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$
Let,
The side opposite to angle θ =AC = k√3
The side adjacent to angle θ =AB = 1k
where k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (1k)² + (k√3)² = (BC)²
⇒ (BC)² = 1 k² +3 k²
⇒ (BC)² = 4 k²
⇒ BC =√2 k²
⇒ BC =±2k
But side BC can’t be negative. So, BC = 2k
Now, we will find the sin θ and cos θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AC = k√3
and Hypotenuse = BC = 2k
So,$\sin \theta=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$
Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = AB =1k
Hypotenuse = BC =2k
So, $\cos \theta=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$

---

Question 18 A 

If $\cot \theta=\frac{21}{20}$, then find the values of cos θ and sin θ.

 

Sol :

We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{21}{20} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{21}{20} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{21}{20}$
Let,
Side adjacent to angle θ =AB = 21k
The side opposite to angle θ =BC = 20k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (21k)² + (20k)² = (AC)²
⇒ (AC)² = 441 k² +400 k²
⇒ (AC)² = 841 k²
⇒ AC =√841 k²
⇒ AC =±29k
But side AC can’t be negative. So, AC = 29k
Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 20k
and Hypotenuse = AC = 29k
So, $\sin \theta=\frac{B C}{A C}=\frac{20 k}{29 k}=\frac{20}{29}$
Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = AB =21k
Hypotenuse = AC =29k
So, $\cos \theta=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{21 \mathrm{k}}{29 \mathrm{k}}=\frac{21}{29}$

---

Question 18 B 

If 15 cot A=18, find sin A and sec A.

Sol :
Given: 15 cot A = 8
$\Rightarrow \cot A=\frac{8}{15}$

And we know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \mathrm{A}=\frac{8}{15} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{8}{15} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{8}{15}$
Let,
Side adjacent to angle A =AB = 8k
The side opposite to angle A =BC = 15k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (8k)² + (15k)² = (AC)²
⇒ (AC)² = 64 k² +225 k²
⇒ (AC)² = 289 k²
⇒ AC =√289 k²
⇒ AC =±17k
But side AC can’t be negative. So, AC = 17k
Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 15k
and Hypotenuse = AC = 17k
So, $\sin \theta=\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}$
Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = AB =8
Hypotenuse = AC =17
So, $\cos \theta=\frac{A B}{A C}=\frac{8 k}{17 k}=\frac{8}{17}$
$\therefore \sec \theta=\frac{1}{\cos \theta}$

$=\frac{1}{\frac{8}{17}}$

$=\frac{17}{8}$

---

Question 19 

If sin θ = cos θ and 0° < θ <90°, then find the values of sin θ and cos θ.

Sol :
Given: sinθ = cosθ
$\Rightarrow \frac{\sin \theta}{\cos \theta}=1$
⇒ tan θ = 1

$\tan \theta=\frac{1}{1} \Rightarrow \frac{P}{B}=\frac{1}{1} \Rightarrow \frac{A B}{B C}=\frac{1}{1}$
Let,
Side opposite to angle θ = AB =1k
The side adjacent to angle θ = BC =1k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (1k)² + (1k)² = (AC)²
⇒ (AC)² = 1k² +1k²
⇒ (AC)² = 2k²
⇒ AC =√2k²
⇒ AC =k√2
So, AC = k√2
Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AB= 1k
and Hypotenuse = AC = k√2
So, $\sin \theta=\frac{A B}{A C}=\frac{1 k}{k \sqrt{2}}=\frac{1}{\sqrt{2}}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = BC =1k
Hypotenuse = AC =k√2
So, $\cos \theta=\frac{B C}{A C}=\frac{1 k}{k \sqrt{2}}=\frac{1}{\sqrt{2}}$

---

Question 20 

If $\sin \theta=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$, then find the values of cos θ and $\frac{1}{\tan \theta}$.

 

Sol :
$\sin \theta=\frac{x^{2}-y^{2}}{x^{2}+v^{2}}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or, $\sin \theta=\frac{\text { Perpendicular }}{\text { hypotenuse }}$

$\sin \theta=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ 

$\Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{\mathrm{x}^{2}+\mathrm{y}^{2}}$ 

$ \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{\mathrm{x}^{2}+\mathrm{y}^{2}}$

Let,
Side opposite to angle θ = AB = x² – y²
and Hypotenuse = AC = x² + y²
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (x² – y² )² + (BC)² = (x² + y² )²
⇒ (BC)² = (x² + y² )² – (x² – y² )²
Using the identity, a² – b² = (a+b)(a – b)
⇒ (BC)² = [(x² + y² + x² – y² )][ x² + y² –( x² – y²)]
⇒ (BC)² = (2x²)(2y²)
⇒ (BC)² = (4x²y²)
⇒ BC =√4x²y²
⇒ BC = ±2xy
⇒ BC = 2xy [taking positive square root since, side cannot be negative]
$\therefore \cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{\text { BC }}{\text { AC }}=\frac{2 x y}{x^{2}+y^{2}}$
and $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}}$
So, $\frac{1}{\tan \theta}=\frac{1}{\frac{x^{2}-y^{2}}{2 x y}}=\frac{2 x y}{x^{2}-y^{2}}$

---

Question 21 

Iftan $\theta=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}}$, then find the values of sin θ and cos θ.

 

Sol :
Given: $\tan \theta=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}}$

We know that,
$\tan \theta=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\sqrt{\mathrm{m}^{2}-\mathrm{n}^{2}}}{\mathrm{n}}$
Let,
AB = √(m² – n²) and BC = n
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (√(m² – n²))² + (n)² = (AC )²
⇒ m² – n² + n² = (AC )²
⇒ (AC)² = (m²)
⇒ AC =√ m²
⇒ AC = ±m
⇒ AC = m [taking positive square root since, side cannot be negative]
Now, we have to find the value of cos θ and sin θ
We, know that
$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
$=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{n}}{\mathrm{m}}$
and
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$=\frac{A B}{A C}=\frac{\sqrt{m^{2}-n^{2}}}{m}$

---

Question 22 A 

If sec θ = 2, then find the values of other t–ratios of angle θ.

 

Sol :
Given: sec θ = 2

We know that,
$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$
$\operatorname{Sec} \theta=\frac{2}{1} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{2}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{2}{1}$
Let,
BC = 1k and AC = 2k
where, k is any positive integer.
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (AB)² + (1k)² = (2k )²
⇒ (AB)² + k² = 4k²
⇒ (AB)² = 4k² – k²
⇒ (AB)² = 3k²
⇒ AB = k√3
Now, we have to find the value of other trigonometric ratios.
We, know that
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$=\frac{A B}{A C}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}$
$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
$=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{1 \mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}$

$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{1 \mathrm{k}}=\frac{\sqrt{3}}{1}=\sqrt{3}$
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$
$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\sqrt{3}}$

---

 

Question 22 B 

Given $\sec \theta=\frac{13}{12}$ calculate all other trigonometric ratios.

 

Sol :

Given: $\sec \theta=\frac{13}{12}$

We know that,
$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$
$\sec \theta=\frac{13}{12} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{13}{12} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13}{12}$
Let,
BC = 12k and AC = 13k
where, k is any positive integer.
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (AB)² + (12k)² = (13k )²
⇒ (AB)² + 144k² = 169k²
⇒ (AB)² = 169k² – 144k²
⇒ (AB)² = 25k²
⇒ AB = √25k²
⇒ AB =±5k [taking positive square root since, side cannot be negative]
Now, we have to find the value of other trigonometric ratios.
We, know that
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$=\frac{A B}{A C}=\frac{5 k}{13 k}=\frac{5}{13}$
$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
$=\frac{B C}{A C}=\frac{12 k}{13 k}=\frac{12}{13}$

$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$=\frac{A B}{B C}=\frac{5 k}{12 k}=\frac{5}{12}$

$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{5}{13}}=\frac{13}{5}$

$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{5}{12}}=\frac{12}{5}$

---

Question 23 

If $\operatorname{cosec} \theta=\sqrt{10}$, then find the values of other t–ratios of angle θ.

 

Sol :

Given: cosec θ = √10

We know that,
$\operatorname{cosec} \theta=\frac{\text { hypotenuse }}{\text { perpendicular }}$
$\operatorname{cosec} \theta=\frac{\sqrt{10}}{1} \Rightarrow \frac{\mathrm{H}}{\mathrm{P}}=\frac{\sqrt{10}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{10}}{1}$
Let,
AB = 1k and AC = k√10
where, k is any positive integer.
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (1k )²+ (BC)² = (k√10)²
⇒ (BC)² = 10k² – k²
⇒ (BC)² = 9k²
⇒ BC = √9k²
⇒ BC =±3k [taking positive square root since, side cannot be negative]
Now, we have to find the value of other trigonometric ratios.
We, know that
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1 \mathrm{k}}{\mathrm{k} \sqrt{10}}=\frac{1}{\sqrt{10}}$

$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
$=\frac{B C}{A C}=\frac{3 k}{k \sqrt{10}}=\frac{3}{\sqrt{10}}$

$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1 \mathrm{k}}{3 \mathrm{k}}=\frac{1}{3}$

$\sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{3}{\sqrt{10}}}=\frac{\sqrt{10}}{3}$

$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{1}{3}}=3$

 

---

Question 24 A 

If $\tan \mathrm{A}=\frac{\sqrt{3}}{2}$ then find the values of sin A + cos A.

 

Sol :

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Given: $\tan \mathrm{A}=\frac{\sqrt{3}}{2}$
$\Rightarrow \tan \mathrm{A}=\frac{\sqrt{3}}{2}$
$\tan \mathrm{A}=\frac{\sqrt{3}}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{2} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\sqrt{3}}{2}$
Let,
Side opposite to angle A =BC = k√3
Side adjacent to angle A =AB = 2k
where, k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (2k)² + (√3k)² = (AC)²
⇒ (AC)² = 4 k² +3 k²
⇒ (AC)² = 7 k²
⇒ AC =√7 k²
⇒ AC =k√7
So, AC = k√7

Now, we will find the sin A and cos A
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle A = BC = k√3
and Hypotenuse = AC = k√7
So, $\sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{k} \sqrt{3}}{\mathrm{k} \sqrt{7}}=\frac{\sqrt{3}}{\sqrt{7}}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle A = AB =2k
Hypotenuse = AC = k√7
So, $\cos \mathrm{A}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{2 \mathrm{k}}{\mathrm{k} \sqrt{7}}=\frac{2}{\sqrt{7}}$
Now, we have to find sin A +cos A

Putting values of sin A and cos A, we get
$\sin \mathrm{A}+\cos \mathrm{A}=\frac{\sqrt{3}}{\sqrt{7}}+\frac{2}{\sqrt{7}}=\frac{\sqrt{3}+2}{\sqrt{7}}$

---

Question 24 B 

If $\sin \theta=\sqrt{3} \cos \theta$ find the value of cos θ – sin θ.

 

Sol :

Given: sin θ =√3cos θ
$\Rightarrow \frac{\sin \theta}{\cos \theta}=\sqrt{3}$
⇒ tan θ = √3

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Given: tan θ = √3
$\Rightarrow \tan \theta=\frac{\sqrt{3}}{1}$
$\tan \theta=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$
Let,
Side opposite to angle θ =AC = √3k
Side adjacent to angle θ =AB = 1k
where k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (1k)² + (√3k)² = (BC)²
⇒ (BC)² = 1 k² +3 k²
⇒ (BC)² = 4 k²
⇒ BC =√2 k²
⇒ BC =±2k
But side BC can’t be negative. So, BC = 2k

Now, we will find the sin B and cos B
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AC = k√3
and Hypotenuse = BC = 2k
So, $\sin \theta=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = AB =1k
Hypotenuse = BC =2k
So, $\cos \theta=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$
Now, we have to find the value of cos θ – sin θ
Putting the values of sin θ and cos θ, we get
$\cos \theta-\sin \theta=\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2}$

---

Question 24 C 

If $\tan \theta=\frac{8}{15}$, find the value of 1+ cos² θ.

 

Sol :

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \theta=\frac{8}{15} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{8}{15} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{8}{15}$
Let,
Side opposite to angle θ =AB = 8k
Side adjacent to angle θ =BC = 15k
where, k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (8k)² + (15k)² = (AC)²
⇒ (AC)² = 64k²+225k²
⇒ (AC)² = 289 k²
⇒ AC =√289 k²
⇒ AC =±17k
But side AC can’t be negative. So, AC = 17k
Now, we will find the cos θ

We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = BC = 15k
and Hypotenuse = AC = 17k
So, $\cos \theta=\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}$
Now, we have to find the value of 1+ cos² θ

Putting the value of cos θ, we get
$1+\cos ^{2} \theta=1+\left(\frac{15}{17}\right)^{2}$

$=1+\frac{225}{289}$

$=\frac{289+225}{289}$

$=\frac{514}{289}$

---

Question 25 

If $\cot \theta=\frac{7}{8}$, evaluate

(i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
(ii) $\cot ^{2} \theta$

 

Sol :

Given: $\cot \theta=\frac{7}{8}$
We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{7}{8} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{7}{8} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{7}{8}$
Let,
Side adjacent to angle θ =AB = 7k
Side opposite to angle θ =BC = 8k
where, k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (7k)² + (8k)² = (AC)²
⇒ (AC)² = 49 k² +69 k²
⇒ (AC)² = 113 k²
⇒ AC =√113 k²
⇒ AC =k√113
$\therefore \sin \theta=\frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{8 \mathrm{k}}{\mathrm{k} \sqrt{113}}=\frac{8}{\sqrt{113}}$

and $\cos \theta=\frac{B}{H}=\frac{A B}{A C}=\frac{7 k}{k \sqrt{113}}=\frac{7}{\sqrt{113}}$

(i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
We know that,
(a+b)(a – b) = (a² – b²)
So, using this identity, we get
$=\frac{(1)^{2}-(\sin \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$

$=\frac{1-\sin ^{2} \theta}{1-\cos ^{2} \theta}$

$=\frac{1-\left(\frac{8}{\sqrt{113}}\right)^{2}}{1-\left(\frac{7}{\sqrt{113}}\right)^{2}}$

$=\frac{1-\frac{64}{113}}{1-\frac{49}{113}}$

$=\frac{\frac{113-64}{113}}{\frac{113-49}{113}}$

$=\frac{49}{64}$

(ii) cot² θ
Given $\cot \theta=\frac{7}{8}$

$=\left(\frac{7}{8}\right)^{2}$

$=\frac{49}{64}$

---

Question 26 A 

If 3 cot A = 4, check whether $\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$=cos²A–sin² A or not.

 

Sol :
Given: 3cot A = 4
$\Rightarrow \cot A=\frac{4}{3}$

We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot A=\frac{4}{3} \Rightarrow \frac{B}{P}=\frac{4}{3} \Rightarrow \frac{A B}{B C}=\frac{4}{3}$
Let,
Side adjacent to angle A =AB = 4k
The side opposite to angle A =BC = 3k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (4k)² + (3k)² = (AC)²
⇒ (AC)² = 16 k² + 9 k²
⇒ (AC)² = 25 k²
⇒ AC =√25k²
⇒ AC = ±5k [taking positive square root since, side cannot be negative]
$\therefore \tan \mathrm{A}=\frac{1}{\cot \mathrm{A}}=\frac{1}{\frac{4}{3}}=\frac{3}{4}$
$\sin \mathrm{A}=\frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3 \mathrm{k}}{5 \mathrm{k}}=\frac{3}{5}$
and $\cos A=\frac{B}{H}=\frac{A B}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

Now, $\mathrm{LHS}=\frac{1-\tan ^{2} \mathrm{A}}{1+\tan ^{2} \mathrm{A}}=\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}$

$=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$

$=\frac{\frac{16-9}{16}}{\frac{16+9}{16}}$

$=\frac{7}{25}$ …(i)

And RHS = cos² A – sin² A

$=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}$

$=\frac{16}{25}-\frac{9}{25}$

$=\frac{7}{25}$ …(ii)

From Eqs. (i) and (ii) LHS =RHS
Hence Proved

---

Question 26 B 

In a right triangle ABC, right angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.

 

Sol :
tan A = 1
As we know
$\tan \theta=\frac{\text { perpedicular }}{\text { base }}$
Now construct a right angle triangle right angled at B such that
∠ BAC = θ
Hence perpendicular = BC = 1 and base = AB = 1

By Pythagoras theorem,
AC² = AB² + BC²
⇒ AC² = (1)² + (1)²
⇒ AC² = 2
⇒ AC = 
As,

$\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}$
 and $\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
⇒ $\sin \theta=\frac{1}{\sqrt{2}}$ and $\cos \theta=\frac{1}{\sqrt{2}}$
Hence,
2 sin A cos A=$2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
⇒ 2 sin A cos A=$2 \times \frac{1}{2}$
⇒ 2 sin A cos A=1
= R.H.S
Hence proved.

---

Question 27 

If 4sin² θ =3 and 0^o < θ <90^o, find the value of 1 + cos θ.

 

Sol :
4sin² θ =3
$\Rightarrow \sin ^{2} \theta=\frac{3}{4}$
$\Rightarrow \sin \theta=\pm \frac{\sqrt{3}}{2}$

But it is given 0^o< θ <90^o
So, $\sin \theta=\frac{\sqrt{3}}{2}$
$\sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{\sqrt{3}}{2}$

Let, P =k√3 and H =2k
In right angled ∆ABC, we have
B² + P² = H²
⇒ B² + (k√3)² = (2k)²
⇒ B² + 3k² = 4k²
⇒ B² = 4k² – 3k²
⇒ B² = k²
⇒ B = ±k
⇒ B = k [taking positive square root since, side cannot be negative]
$\therefore \cos \theta=\frac{B}{H}=\frac{k}{2 k}=\frac{1}{2}$
So, $1+\cos \theta=1+\frac{1}{2}=\frac{2+1}{2}=\frac{3}{2}$

---

Question 28 

If $\tan \theta=\frac{p}{q}$find the value of $\frac{p \sin \theta-q \cos \theta}{p \sin \theta+q \cos \theta}$.

 

Sol :
Given:$\tan \theta=\frac{p}{q}$

Now, $\frac{p \sin \theta-q \cos \theta}{p \sin \theta+q \cos \theta}$
$\Rightarrow \frac{\cos \theta\left(\frac{p \sin \theta}{\cos \theta}-q\right)}{\cos \theta\left(\frac{p \sin \theta}{\cos \theta}+q\right)}$

$\Rightarrow \frac{p \tan \theta-q}{p \tan \theta+q}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$

$\Rightarrow \frac{p\left(\frac{p}{q}\right)-q}{p\left(\frac{p}{q}\right)+q}$
$\Rightarrow \frac{\frac{p^{2}-q^{2}}{q}}{\frac{p^{2}+q^{2}}{q}}$
$\Rightarrow \frac{p^{2}-q^{2}}{p^{2}+q^{2}}$
$\therefore \frac{p \sin \theta-q \cos \theta}{p \sin \theta+q \cos \theta}=\frac{p^{2}-q^{2}}{p^{2}+q^{2}}$

---

Question 29 

If 13 cos θ = 5, $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}$.

 

Sol :
Given: 13 cosθ = 5
$\Rightarrow \cos \theta=\frac{5}{13}$

We know that,
$\cos \theta=\frac{\text { Base }}{\text { hypotenuse }}$
$\cos \theta=\frac{5}{13} \Rightarrow \frac{B}{H}=\frac{5}{13}$

Let AB =5k and BC = 13k
In right angled ∆ABC, we have
B² + P² = H²
⇒ (5k)² + P² = (13k)²
⇒ P² + 25k² = 169k²
⇒ P² = 169k² – 25k²
⇒ P² = 144k²
⇒ P =√144k²
⇒ P = ±12k
⇒ P = 12k [taking positive square root since, side cannot be negative]
$\therefore \sin \theta=\frac{P}{H}=\frac{12}{13}$

Now, $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}$
$=\frac{\frac{12}{13}+\frac{5}{13}}{\frac{12}{13}-\frac{5}{13}}$
$=\frac{17}{7}$

---

Question 30 

If $\sec \theta=\frac{13}{5}$, show that $\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}=3$.

Sol :
Given: 

$\sec \theta=\frac{13}{5}$

We know that,
$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$
$\operatorname{Sec} \theta=\frac{13}{5} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{13}{5} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13}{5}$
Let,
BC = 5k and AC = 13k
where, k is any positive integer.
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (AB)² + (5k)² = (13k )²
⇒ (AB)² + 25k² = 169k²
⇒ (AB)² = 169k² – 25k²
⇒ (AB)² = 144k²
⇒ AB = √144k²
⇒ AB =±12k [taking positive square root since, side cannot be negative]
Now, we have to find the value of other trigonometric ratios.
We, know that
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$=\frac{A B}{A C}=\frac{12 k}{13 k}=\frac{12}{13}$

$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
$=\frac{B C}{A C}=\frac{5 k}{13 k}=\frac{5}{13}$

Now, LHS $=\frac{2 \sin \theta-3 \cos \theta}{4 \sin \theta-9 \cos \theta}$
$=\frac{2\left(\frac{12}{13}\right)-3\left(\frac{5}{13}\right)}{4\left(\frac{12}{13}\right)-9\left(\frac{5}{13}\right)}$

$=\frac{24-15}{48-45}$

$=\frac{9}{3}$

=3 =RHS
Hence Proved

---

Question 31 

If 2 tan θ = 1, find the value of $\frac{3 \cos \theta+\sin \theta}{2 \cos \theta-\sin \theta}$.

 

Sol :
Given: 2 tan θ = 1
$\Rightarrow \tan \theta=\frac{1}{2}$

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \theta=\frac{1}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{1}{2} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{2}$
Let,
Side opposite to angle θ =AB = 1k
Side adjacent to angle θ =BC = 2k
where, k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (k)² + (2k)² = (AC)²
⇒ (AC)² = k²+4k²
⇒ (AC)² = 5k²
⇒ AC =√5k²
⇒ AC =±k√5
But side AC can’t be negative. So, AC = k√5

Now, we will find the sin θ and cos θ

We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = BC = 2k
and Hypotenuse = AC = k√5
So,$\cos \theta=\frac{B C}{A C}=\frac{2 k}{k \sqrt{5}}=\frac{2}{\sqrt{5}}$
And $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ =AB = 1k
And Hypotenuse =AC = k√5
So, $\sin \theta=\frac{A B}{A C}=\frac{1 k}{k \sqrt{5}}=\frac{1}{\sqrt{5}}$
Now, $\frac{3 \cos \theta+\sin \theta}{2 \cos \theta-\sin \theta}$
$=\frac{3\left(\frac{2}{\sqrt{5}}\right)+\frac{1}{\sqrt{5}}}{2\left(\frac{2}{\sqrt{5}}\right)-\frac{1}{\sqrt{5}}}$

$=\frac{6+1}{4-1}$

$=\frac{7}{3}$

---

Question 32 

If 5 tan α = 4, show that $\frac{5 \sin \alpha-3 \cos \alpha}{5 \sin \alpha+2 \cos \alpha}=\frac{1}{6}$.

 

Sol :
Given: 5 tan = 4
$\Rightarrow \tan \alpha=\frac{4}{5}$

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \alpha=\frac{4}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{4}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{4}{5}$

Let,
The side opposite to angle α =AB = 4k
The side adjacent to angle α =BC = 5k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (4k)² + (5k)² = (AC)²
⇒ (AC)² = 16k²+25k²
⇒ (AC)² = 41k²
⇒ AC =√41k²
⇒ AC =±k√41
But side AC can’t be negative. So, AC = k√41
Now, we will find the sin α and cos α
We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle α = BC = 5k
and Hypotenuse = AC = k√41
So, $\cos \alpha=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5 \mathrm{k}}{\mathrm{k} \sqrt{41}}=\frac{5}{\sqrt{41}}$
And $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle α =AB = 4k
And Hypotenuse =AC = k√5
So, $\sin \alpha=\frac{A B}{A C}=\frac{4 k}{k \sqrt{4} 1}=\frac{4}{\sqrt{41}}$

Now, LHS $=\frac{5 \sin \alpha-3 \cos \alpha}{5 \sin \alpha+2 \cos \alpha}$
$=\frac{5\left(\frac{4}{\sqrt{41}}\right)-3\left(\frac{5}{\sqrt{41}}\right)}{5\left(\frac{4}{\sqrt{41}}\right)+2\left(\frac{5}{\sqrt{41}}\right)}$
$=\frac{20-15}{20+10}$

$=\frac{5}{30}$

$=\frac{1}{6}$

 = RHS
Hence Proved

---

Question 33 

If $\cot \theta=\frac{3}{4}$prove that $\sqrt{\frac{\sec \theta+\operatorname{cosec} \theta}{\sec \theta-\operatorname{cosec} \theta}}=\sqrt{7}$.

 

Sol :

We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{3}{4} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{3}{4}$
Let,
Side adjacent to angle θ =AB = 3k
The side opposite to angle θ =BC = 4k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (4k)² = (AC)²
⇒ (AC)² = 9k² +16k²
⇒ (AC)² = 25k²
⇒ AC =√25k²
⇒ AC =±5k
But side AC can’t be negative. So, AC = 5k

Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 4k
and Hypotenuse = AC = 5k
So, $\sin \theta=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle θ = AB =3k
Hypotenuse = AC =5k
So, $\cos \theta=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$
$\therefore \sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{3}{5}}=\frac{5}{3}$
And
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{4}{5}}=\frac{5}{4}$

Now, LHS$=\sqrt{\frac{\sec \theta+\operatorname{cosec} \theta}{\sec \theta-\operatorname{cosec} \theta}}$

$=\sqrt{\frac{\left(\frac{5}{3}\right)+\left(\frac{5}{4}\right)}{\left(\frac{5}{3}\right)-\left(\frac{5}{4}\right)}}$

$=\sqrt{\frac{\left(\frac{20+15}{12}\right)}{\left(\frac{20-15}{12}\right)}}$

$=\sqrt{\frac{35}{5}}$

$=\sqrt{7}$ = RHS
Hence Proved

---

Question 34

If $\cot \theta=\frac{1}{\sqrt{3}}$ verify that: $\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}=\frac{3}{5}$.

 

Sol :

We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1}{\sqrt{3}}$

Let,
Side adjacent to angle θ =AB = 1k
Side opposite to angle θ =AC = k√3
where, k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (1k)² + (√3k)² = (BC)²
⇒ (BC)² = 1 k² +3 k²
⇒ (BC)² = 4 k²
⇒ BC =√2 k²
⇒ BC =±2k
But side BC can’t be negative. So, BC = 2k
Now, we will find the sin θ and cos θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = AC = k√3
and Hypotenuse = BC = 2k
So$\sin \theta=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = AB =1k
Hypotenuse = BC =2k
So, $\cos \theta=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$

Now, LHS $=\frac{1-\cos ^{2} \theta}{2-\sin ^{2} \theta}$
$=\frac{1-\left(\frac{1}{2}\right)^{2}}{2-\left(\frac{\sqrt{3}}{2}\right)^{2}}$
$=\frac{1-\frac{1}{4}}{2-\frac{3}{4}}$
$=\frac{\frac{4-1}{4}}{\frac{8-3}{4}}$
$=\frac{3}{5}$ = RHS
Hence Proved

---

Question 35 

If $\tan \theta=\frac{x}{y}$ find the value of x sin θ + y cos θ.

 

Sol :

We know that,

$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
$\tan \theta=\frac{x}{y} \Rightarrow \frac{P}{B}=\frac{x}{y} \Rightarrow \frac{A B}{B C}=\frac{x}{y}$
Let,
Side opposite to angle θ =AB = x
Side adjacent to angle θ =BC = y
where, k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (x)² + (y)² = (AC)²
⇒ (AC)² = x²+y²
⇒ AC =√( x²+y²)
Now, we will find the sin θ and cos θ
We know that
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = BC = y
and Hypotenuse = AC = √( x²+y²)
So,$\cos \theta=\frac{B C}{A C}=\frac{y}{\sqrt{x^{2}+y^{2}}}$
And $\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ =AB = x
And Hypotenuse =AC = √( x²+y²)
So, $\sin \theta=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}$
Now, x sin θ +y cos θ
$=\mathrm{x}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}\right)+\mathrm{y}\left(\frac{\mathrm{y}}{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}}\right)$
$=\frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}}}$
= √( x²+y²)

---

Question 36 

If $\sin \theta=\frac{3}{5}$, find the value of tan²θ + sinθ cosθ + cotθ.

 

Sol :

Given: $\sin \theta=\frac{3}{5}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypollyuse }}$
$\sin \theta=\frac{3}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{5}$
Let,
Perpendicular =AB =3k
and Hypotenuse =AC =5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
In right angled ∆ ABC, we have
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (BC)² = (5k)²
⇒ 9k² + (BC)² = 25k²
⇒ (BC)² = 25 k² –9k²
⇒ (BC)² = 16k²
⇒ BC =√16k²
⇒ BC =±4k
But side BC can’t be negative. So, BC = 4k
Now, we have to find the value of cos θ and tan θ
We know that,
$\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
The side adjacent to angle θ or base = BC =4k
Hypotenuse = AC =5k
So,$\cos \theta=\frac{4 k}{5 k}=\frac{4}{5}$
Now,
We know that,
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Perpendicular = AB =3k
Base = BC =4k
So, $\tan \theta=\frac{3 \mathrm{k}}{4 \mathrm{k}}=\frac{3}{4}$
$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$

Now, tan² θ + sin θ cos θ + cot θ

$=\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{5}\right)\left(\frac{4}{5}\right)+\left(\frac{4}{3}\right)$

$=\left(\frac{9}{16}\right)+\left(\frac{13}{25}\right)+\left(\frac{4}{3}\right)$

$=\frac{675+576+1600}{16 \times 25 \times 3}$

$=\frac{2851}{1200}$

---

Question 37 

If 4cot θ = 3, show that $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}=7$.

 

Sol :
Given: $\cot \theta=\frac{3}{4}$

We know that,

$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot \theta=\frac{3}{4} \Rightarrow \frac{\mathrm{B}}{\mathrm{P}}=\frac{3}{4} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{3}{4}$
Let,
Side adjacent to angle θ =AB = 3k
The side opposite to angle θ =BC = 4k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (4k)² = (AC)²
⇒ (AC)² = 9k² +16k²
⇒ (AC)² = 25k²
⇒ AC =√25k²
⇒ AC =±5k
But side AC can’t be negative. So, AC = 5k

Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle θ = BC = 4k
and Hypotenuse = AC = 5k
So, $\sin \theta=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle θ = AB =3k
Hypotenuse = AC =5k
So, $\cos \theta=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$

Now, LHS $=\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}$

$=\frac{\frac{4}{5}+\frac{3}{5}}{\frac{4}{5}-\frac{3}{5}}$

$=\frac{7}{1}$

= 7 = RHS
Hence Proved

---

Question 38 

If $\sin \theta=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$, prove that $\mathrm{m} \sin \theta+\mathrm{n} \cos \theta=\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}$

 

Sol :
Given: $\sin \theta=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$

Let,
Perpendicular =AB =m
and Hypotenuse =AC =√(m² + n²)
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
In right angled ∆ ABC, we have
⇒ (AB)² + (BC)² = (AC)²
⇒ (m)² + (BC)² = (√(m² + n²))²
⇒ m² + (BC)² = m² + n²
⇒ (BC)² = m² + n² – m²
⇒ (BC)² = n²
⇒ BC =√n²
⇒ BC =±n
But side BC can’t be negative. So, BC = n
Now, we have to find the value of cos θ and tan θ

We know that,
$\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
Side adjacent to angle θ or base = BC =n
Hypotenuse = AC =√(m² + n²)
So, $\cos \theta=\frac{n}{\sqrt{m^{2}+n^{2}}}$

Now, LHS = m sin θ +n cosθ
$=\mathrm{m}\left(\frac{\mathrm{m}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}\right)+\mathrm{n}\left(\frac{\mathrm{n}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}\right)$
$=\frac{\mathrm{m}^{2}+\mathrm{n}^{2}}{\sqrt{\mathrm{m}^{2}+\mathrm{n}^{2}}}$
=√(m² + n²) = RHS
Hence Proved

---

Question 39 

If $\cos \alpha=\frac{12}{13}$ show that $\sin \alpha(1-\tan \alpha)=\frac{35}{156}$

 

Sol :

We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Or $\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}$
$\cos \alpha=\frac{12}{13} \Rightarrow \frac{\mathrm{B}}{\mathrm{H}}=\frac{12}{13} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{12}{13}$

Let,
Base =BC = 12k
Hypotenuse =AC = 13k
Where, k ia any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (AB)² + (12k)² = (13k)²
⇒ (AB)² + 144k² = 169k²
⇒ (AB)² = 169 k² –144 k²
⇒ (AB)² = 25 k²
⇒ AB =√25 k²
⇒ AB =±5k
But side AB can’t be negative. So, AB = 5k
Now, we have to find sin α and tan α

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle α = AB =5k
And Hypotenuse = AC =13k
So, $\sin \alpha=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Side opposite to angle α = AB =5k
Side adjacent to angle α = BC =12k
So, $\tan \alpha=\frac{5 \mathrm{k}}{12 \mathrm{k}}=\frac{5}{12}$
Now, LHS = sin α (1 – tan α)

$=\frac{5}{13}\left(1-\frac{5}{12}\right)$

$=\frac{5}{13}\left(\frac{12-5}{12}\right)$

$=\frac{35}{156}$ = RHS
Hence Proved

---

Question 40 

If $\mathrm{q} \cos \theta=\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}$, prove that q sin θ = p.

 

Sol :
Given : q cos θ = √(q² – p²)
$\Rightarrow \cos \theta=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}}$

We know that,

$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Or $\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}$
$\cos \theta=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}} \Rightarrow \frac{\mathrm{B}}{\mathrm{H}}=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\sqrt{\mathrm{q}^{2}-\mathrm{p}^{2}}}{\mathrm{q}}$

Let,
Base =BC = √(q² – p²)
Hypotenuse =AC = q
Where, k ia any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AB)² + (BC)² = (AC)²
⇒ (AB)² + (√(q² – p²))² = (q)²
⇒ (AB)² + (q² – p²) = q²
⇒ (AB)² = q² – q² + p²)
⇒ (AB)² = p²
⇒ AB =√p²
⇒ AB =±p
But side AB can’t be negative. So, AB = p
Now, we have to find sin θ

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
The side opposite to angle θ = AB =p
And Hypotenuse = AC =q
So, $\sin \theta=\left(\frac{p}{q}\right)$
Now, LHS = q sin θ
$=\mathrm{q}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)$
= q = RHS
Hence Proved

---

Question 41 

If $\sin \theta=\frac{3}{5}$, show that : 

$\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}=-\frac{1}{5}$

 

Sol :
Given: $\sin \theta=\frac{3}{5}$

We know that,

$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \theta=\frac{3}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{3}{5} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{5}$
Let,
Perpendicular =AB =3k
and Hypotenuse =AC =5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
In right angled ∆ ABC, we have
⇒ (AB)² + (BC)² = (AC)²
⇒ (3k)² + (BC)² = (5k)²
⇒ 9k² + (BC)² = 25k²
⇒ (BC)² = 25 k² –9k²
⇒ (BC)² = 16k²
⇒ BC =√16k²
⇒ BC =±4k
But side BC can’t be negative. So, BC = 4k
Now, we have to find the value of cos θ and tan θ

We know that,
$\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
The side adjacent to angle θ or base = BC =4k
Hypotenuse = AC =5k
So, $\cos \theta=\frac{4 k}{5 k}=\frac{4}{5}$
Now,
We know that,
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Perpendicular = AB =3k
Base = BC =4k
So, $\tan \theta=\frac{3 \mathrm{k}}{4 \mathrm{k}}=\frac{3}{4}$
$\cot \theta=\frac{1}{\tan \theta}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$
Now, LHS $=\frac{\cos \theta-\frac{1}{\tan \theta}}{2 \cot \theta}$
$=\frac{\left(\frac{4}{5}\right)-\left(\frac{1}{\frac{3}{4}}\right)}{2\left(\frac{4}{3}\right)}$
$=\frac{\left(\frac{4}{5}\right)-\left(\frac{4}{3}\right)}{\left(\frac{8}{3}\right)}$
$=\frac{\frac{12-20}{15}}{\left(\frac{8}{3}\right)}$
$=\frac{\left(-\frac{8}{15}\right)}{\left(\frac{8}{3}\right)}$
$=-\frac{1}{5}$ = RHS
Hence Proved

---

Question 42 A 

Find the value of

cos A sin B + sin A. cos B, if sin A= 4/5 and cos B = 12/13.
Sol :

Given: $\sin A=\frac{4}{5}$ and $\cos B=\frac{12}{13}$
To find: cos A sin B + sin A cos B
As, we have the value of sin A and cos B but we don’t have the value of cos A and sin B
So, First we find the value of cos A and sin B
$\sin A=\frac{4}{5}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

$\sin A=\frac{4}{5} \Rightarrow \frac{P}{H}=\frac{4}{5}$
Let,
Side opposite to angle A = 4k
and Hypotenuse = 5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (P)² + (B)² = (H)²
⇒ (4k)² + (B)² = (5)²
⇒ 16 k² + (B)² = 25 k²
⇒ (B)² = 25 k² –16 k²
⇒ (B)² = 9 k²
⇒ B =√9 k²
⇒ B =±3k [taking positive square root since, side cannot be negative]
So, Base = 3k
Now, we have to find the value of cos A
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle A =3k
Hypotenuse =5k
So, $\cos \mathrm{A}=\frac{3 \mathrm{k}}{5 \mathrm{k}}=\frac{3}{5}$
Now, we have to find the sin B

We know that,

$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
$\cos \mathbf{B}=\frac{12}{13} \Rightarrow \frac{B}{H}=\frac{12}{13}$
Let,
Side adjacent to angle B =12k
Hypotenuse =13k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (B)² + (P)² = (H)²
⇒ (12k)² + (P)² = (13)²
⇒ 144 k² + (P)² = 169 k²
⇒ (P)² = 169 k² –144 k²
⇒ (P)² = 25 k²
⇒ P =√25 k²
⇒ P =±5k [taking positive square root since, side cannot be negative]
So, Perpendicular = 5k
Now, we have to find the value of sin B

We know that,
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \mathrm{B}=\frac{\mathrm{P}}{\mathrm{H}}=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$
Now, cos A sin B + sin A cos B

Putting the values of sin A, sin B cos A and Cos B, we get

$\Rightarrow\left(\frac{3}{5}\right)\left(\frac{5}{13}\right)+\left(\frac{4}{5}\right)\left(\frac{12}{13}\right)$

$\Rightarrow \frac{15+48}{5 \times 13}$

$\Rightarrow \frac{63}{65}$

---

Question 42 B 

Find the value of

sin A. cos B – cos A. sin B, if tan A= √3 and sin B = 1/2.
Sol :

Given: tan A =√3 and $\sin B=\frac{1}{2}$
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \mathrm{B}=\frac{1}{2} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{1}{2}$
Let,
Side opposite to angle θ = 1k
and Hypotenuse = 2k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AC)² + (BC)² = (AB)²
⇒ (1k)² + (BC)² = (2k)²
⇒ k² + (BC)² = 4k²
⇒ (BC)² = 4k² –k²
⇒ (BC)² = 3 k²
⇒ BC =√3k²
⇒ BC =k√3
So, BC = k√3

Now, we have to find the value of cos B
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle B = BC =k√3
Hypotenuse = AB =2k
So, $\cos \mathbf{B}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}$

We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
Or $\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
Given: tan A = √3
$\Rightarrow \tan \mathrm{A}=\frac{\sqrt{3}}{1}$
$\tan \mathrm{A}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$
Let,
The side opposite to angle A =BC = √3k
The side adjacent to angle A =AB = 1k
where k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (1k)² + (√3k)² = (AC)²
⇒ (AC)² = 1 k² +3 k²
⇒ (AC)² = 4 k²
⇒ AC =√2 k²
⇒ AC =±2k
But side AC can’t be negative. So, AC = 2k
Now, we will find the sin A and cos A
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle A = BC = k√3
and Hypotenuse = AC = 2k
So, $\sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
The side adjacent to angle A = AB =1k
Hypotenuse = AC =2k

So, $\cos \mathrm{A}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1 \mathrm{k}}{2 \mathrm{k}}=\frac{1}{2}$
Now, sin A. cos B – cos A. sin B

Putting the values of sin A, sin B cos A and Cos B, we get

$\Rightarrow\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$

$\Rightarrow \frac{3}{4}-\frac{1}{4}$

$\Rightarrow \frac{2}{4}$

$=\frac{1}{2}$

---

Question 42 C 

Find the value of

sin A. cos B + cos A. sin B. if $\tan \mathrm{A}=\frac{1}{\sqrt{3}}$and tan B = √3.
Sol :
Given:

$\tan \mathrm{A}=\frac{1}{\sqrt{3}}$
$\tan \mathrm{A}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{1}{\sqrt{3}}$
Let,
Side opposite to angle A =BC = 1k
Side adjacent to angle A =AB = k√3
where, k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (√3k)² + (1k)² = (AC)²
⇒ (AC)² = 1 k² +3 k²
⇒ (AC)² = 4 k²
⇒ AC =√2 k²
⇒ AC =±2k
But side AC can’t be negative. So, AC = 2k

Now, we will find the sin A and cos A

$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle A = BC = k
and Hypotenuse = AC = 2k
So, $\operatorname{Sin} \mathbf{A}=\frac{B C}{A C}=\frac{1 k}{2 k}=\frac{1}{2}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle A = AB =k√3
Hypotenuse = AC =2k
So, $\cos \mathbf{A}=\frac{A B}{B C}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}$
Now,

Given: tan B = √3
$\Rightarrow \tan \mathrm{B}=\frac{\sqrt{3}}{1}$
$\tan \mathrm{B}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{P}}{\mathrm{B}}=\frac{\sqrt{3}}{1} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\sqrt{3}}{1}$
Let,
Side opposite to angle B =AC = √3k
Side adjacent to angle B =AB = 1k
where, k is any positive integer
Firstly we have to find the value of BC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (AC)² = (BC)²
⇒ (1k)² + (√3k)² = (BC)²
⇒ (BC)² = 1 k² +3 k²
⇒ (BC)² = 4 k²
⇒ BC =√2 k²
⇒ BC =±2k
But side BC can’t be negative. So, BC = 2k

Now, we will find the sin B and cos B
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle B = AC = k√3
and Hypotenuse = BC = 2k
So, $\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$

Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle B = AB =1k
Hypotenuse = BC =2k
So, $\cos \mathbf{B}=\frac{A B}{B C}=\frac{1 k}{2 k}=\frac{1}{2}$
Now, sin A. cos B + cos A. sin B

Putting the values of sin A, sin B cos A and Cos B, we get

$\Rightarrow\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$

$\Rightarrow \frac{1}{4}+\frac{3}{4}$

$\Rightarrow \frac{4}{4}$

=1

---

Question 42 D 

Find the value of

$\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}$, if sin A = $\frac{1}{\sqrt{2}}$and cos $\mathrm{B}=\frac{\sqrt{3}}{2}$
Sol :
Given $: \sin \mathrm{A}=\frac{1}{\sqrt{2}}$ and $\cos \mathrm{B}=\frac{\sqrt{3}}{2}$
$\sin \mathrm{A}=\frac{1}{\sqrt{2}}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \mathrm{A}=\frac{1}{\sqrt{2}} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{1}{\sqrt{2}}$

Let,
Side opposite to angle A = k
and Hypotenuse = k√2
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (P)² + (B)² = (H)²
⇒ (k)² + (B)² = (k√2)²
⇒ k² + (B)² = 2k²
⇒ (B)² = 2k² – k²
⇒ (B)² = k²
⇒ B =√k²
⇒ B =±k [taking positive square root since, side cannot be negative]
So, Base = k
Now, we have to find the value of tan A

We know that,
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
So, $\tan \mathrm{A}=\frac{\mathrm{k}}{\mathrm{k}}=1$

Now, we have to find the tan B
We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
$\cos \mathbf{B}=\frac{\sqrt{3}}{2} \Rightarrow \frac{B}{H}=\frac{\sqrt{3}}{2}$
Let,
Side adjacent to angle B =k√3
Hypotenuse =2k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (B)² + (P)² = (H)²
⇒ (k√3)² + (P)² = (2k)²
⇒ 3k² + (P)² = 4k²
⇒ (P)² = 4k² –3 k²
⇒ (P)² = k²
⇒ P =√k²
⇒ P =±k [taking positive square root since, side cannot be negative]
So, Perpendicular = k
Now, we have to find the value of sin B
We know that,
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$
So, $\tan \mathrm{B}=\frac{\mathrm{k}}{\mathrm{k} \sqrt{3}}=\frac{1}{\sqrt{3}}$

Now, $\frac{\tan A+\tan B}{1-\tan A \tan B}$
$\Rightarrow \frac{(1)+\left(\frac{1}{\sqrt{3}}\right)}{1-(1)\left(\frac{1}{\sqrt{3}}\right)}$

$\Rightarrow \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}$

$\Rightarrow \frac{\sqrt{3}+1}{\sqrt{3}-1}$

Now, multiply and divide by the conjugate of √3 – 1, we get
$\Rightarrow \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\Rightarrow \frac{(\sqrt{3}+1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}$ [∵ (a – b)(a+b) = (a² – b²)]

$\Rightarrow \frac{3+1+2 \sqrt{3}}{3-1}$

$\Rightarrow \frac{4+2 \sqrt{3}}{2}$

⇒ 2+√3

---

Question 42 E

Find the value of

sec A. tan A+tan²A – cosec A, if tan A =2

 

Sol :
Given: tan A = 2 ⇒ tan²A = 4
We know that, sec² A = 1+ tan²A
⇒ sec² A = 1 + 4
⇒ sec² A = 5
⇒ sec A =√5
$\Rightarrow \cos A=\frac{1}{\sqrt{5}}$

Now, we know that tan A$=\frac{\sin A}{\cos A}$
$\Rightarrow 2=\frac{\sin A}{\frac{1}{\sqrt{5}}}$
⇒ 2 =√5 sin A

$\Rightarrow \sin A=\frac{2}{\sqrt{5}}$

$\Rightarrow \operatorname{cosec} A=\frac{\sqrt{5}}{2}$

Now, putting all the values in the given equation, we get
sec A. tan A+tan²A – cosec A

$\Rightarrow(\sqrt{5})(2)+(4)-\left(\frac{\sqrt{5}}{2}\right)$

$\Rightarrow \frac{4 \sqrt{5}+8-\sqrt{5}}{2}$

$\Rightarrow \frac{3 \sqrt{5}+8}{2}$

---

Question 42 F

Find the value of

$\frac{1}{\tan \mathrm{A}}+\frac{\sin \mathrm{A}}{1+\cos \mathrm{A}}$, if cosec A = 2
Sol :
Given: cosec A =2
Now, we have to find $\frac{1}{\tan A}+\frac{\sin A}{1+\cos A}$
First, we simplify the above given trigonometry equation, we get
$\frac{1}{\frac{\sin A}{\cos A}}+\frac{\sin A}{1+\cos A}$
$\Rightarrow \frac{\cos A}{\sin A}+\frac{\sin A}{1+\cos A}$

Taking the LCM, we get
$\Rightarrow \frac{\cos A(1+\cos A)+\sin A(\sin A)}{(\sin A)(1+\cos A)}$
$\Rightarrow \frac{\cos A+\cos ^{2} A+\sin ^{2} A}{\sin A(1+\cos A)}$ [∵ cos²θ +sin² θ = 1]
$\Rightarrow \frac{\cos A+1}{\sin A(1+\cos A)}$
$\Rightarrow \frac{1}{\sin A}$ [∵ cosec θ $=\frac{1}{\sin \theta}$ ]
⇒ cosec A
⇒ 2

---

Question 43 A

If $\sin \mathrm{B}=\frac{1}{2}$, prove that : 3 cos B – 4cos³ B = 0

 

Sol :
Given: $\sin B=\frac{1}{2}$

We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin B=\frac{1}{2} \Rightarrow \frac{P}{H}=\frac{1}{2} \Rightarrow \frac{A B}{A C}=\frac{1}{2}$
Let,
Perpendicular =AB =k
and Hypotenuse =AC =2k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
In right angled ∆ ABC, we have
⇒ (AB)² + (BC)² = (AC)²
⇒ (k)² + (BC)² = (2k)²
⇒ k² + (BC)² = 4k²
⇒ (BC)² = 4k² –k²
⇒ (BC)² = 3k²
⇒ BC =√3k²
⇒ BC =k√3
So, BC = k√3
Now, we have to find the value of cos B

We know that,
$\cos \theta=\frac{\text { base }}{\text { hypotenuse }}$
Side adjacent to angle B or base = BC = k√3
Hypotenuse = AC =2k
So, $\cos \mathrm{B}=\frac{\mathrm{k} \sqrt{3}}{2 \mathrm{k}}=\frac{\sqrt{3}}{2}$
Now, LHS = 3 cos B – 4cos³ B

$\Rightarrow 3\left(\frac{\sqrt{3}}{2}\right)-4\left(\frac{\sqrt{3}}{2}\right)^{3}$

$\Rightarrow \frac{3 \sqrt{3}}{2}-4\left(\frac{3 \sqrt{3}}{8}\right)$

$\Rightarrow \frac{3 \sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}=0$

=RHS
Hence Proved

---

Question 43 B

If $\cos \theta=\frac{\sqrt{3}}{2}$, prove that: 3sinθ – 4sin³θ = 1.

 

Sol :

We know that,

$\cos \theta=\frac{\text { Base }}{\text { hypotenuse }}$
$\cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \frac{\mathrm{B}}{\mathrm{H}}=\frac{\sqrt{3}}{2}$
Let AB =k√3 and BC = 2k
In right angled ∆ABC, we have
B² + P² = H²
⇒ (k√3)² + P² = (2k)²
⇒ P² + 3k² = 4k²
⇒ P² = 4k² – 3k²
⇒ P² = k²
⇒ P =√k²
⇒ P = ±k
⇒ P = k [taking positive square root since, side cannot be negative]
Now,
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\therefore \sin \theta=\frac{P}{H}=\frac{k}{2 k}=\frac{1}{2}$

Now, LHS = 3sin θ – 4sin³ θ

$=3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^{3}$

$\Rightarrow \frac{3}{2}-\frac{4}{8}$

$\Rightarrow \frac{3}{2}-\frac{1}{2}$

$\Rightarrow \frac{3-1}{2}$

$\Rightarrow \frac{2}{2}$

⇒ 1 = RHS
Hence Proved

---

Question 43 C

If $\sec \theta=\frac{5}{4}$, prove that : 

$\frac{\tan \theta}{1+\tan ^{2} \theta}=\frac{\sin \theta}{\sec \theta}$

 

Sol :
Given: $\sec \theta=\frac{5}{4}$

We know that,

$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}$
$\operatorname{Sec} \theta=\frac{5}{4} \Rightarrow \frac{\mathrm{H}}{\mathrm{B}}=\frac{5}{4} \Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{5}{4}$
Let,
BC = 4k and AC = 5k
where, k is any positive integer.
In right angled ∆ABC, we have
(AB)² + (BC)² = (AC)² [by using Pythagoras theorem]
⇒ (AB)² + (4k)² = (5k )²
⇒ (AB)² + 16k² = 25k²
⇒ (AB)² = 25k² – 16k²
⇒ (AB)² = 9k²
⇒ AB = √9k²
⇒ AB =±3k [taking positive square root since, side cannot be negative]
Now, we have to find the value of other trigonometric ratios.
We, know that

$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

$=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$

$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

$=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$

$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

$=\frac{A B}{B C}=\frac{3 k}{4 k}=\frac{3}{4}$

Now, LHS $=\frac{\tan \theta}{1+\tan ^{2} \theta}$

$=\frac{\left(\frac{3}{4}\right)}{1+\left(\frac{3}{4}\right)^{2}}$

$=\frac{\left(\frac{3}{4}\right)}{1+\left(\frac{9}{16}\right)}$

$=\frac{\left(\frac{3}{4}\right)}{\left(\frac{16+9}{16}\right)}$

$=\frac{\left(\frac{3}{4}\right)}{\left(\frac{25}{16}\right)}$

$=\frac{3}{4} \times \frac{16}{25}$

$=\frac{12}{25}$

Now, RHS $=\frac{\sin \theta}{\sec \theta}$

$=\frac{\left(\frac{3}{5}\right)}{\left(\frac{5}{4}\right)}$

$=\frac{3}{5} \times \frac{4}{5}$

$=\frac{12}{25}$

∴ LHS = RHS
Hence Proved

---

Question 43 D

$\cot B=\frac{12}{5}$

, prove that : tan²B – sin² B=sin⁴ B sec² B.

 

Sol :

We know that,
$\cot \theta=\frac{\text { side adjacent to angle } \theta}{\text { side opposite to angle } \theta}$
Or $\cot \theta=\frac{\text { base }}{\text { perpendicular }}$
$\cot B=\frac{12}{5} \Rightarrow \frac{B}{P}=\frac{12}{5} \Rightarrow \frac{A B}{B C}=\frac{12}{5}$
Let,
Side adjacent to angle B =AB = 12k
Side opposite to angle B =BC = 5k
where, k is any positive integer
Firstly we have to find the value of AC.
So, we can find the value of AC with the help of Pythagoras theorem
⇒ (AB)² + (BC)² = (AC)²
⇒ (12k)² + (5k)² = (AC)²
⇒ (AC)² = 144 k² +25 k²
⇒ (AC)² = 169 k²
⇒ AC =√169 k²
⇒ AC =±13k
But side AC can’t be negative. So, AC = 13k

Now, we will find the sin θ
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Side opposite to angle B = BC = 5k
and Hypotenuse = AC = 13k
So, $\sin \mathrm{B}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5 \mathrm{k}}{13 \mathrm{k}}=\frac{5}{13}$
Now, we know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle B = AB =12k
Hypotenuse = AC =13k
So, $\cos \mathrm{B}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12 \mathrm{k}}{13 \mathrm{k}}=\frac{12}{13}$
$\tan \mathrm{B}=\frac{\mathrm{P}}{\mathrm{B}}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}$
$\sec \mathrm{B}=\frac{1}{\cos \mathrm{B}}=\frac{1}{\frac{12}{13}}=\frac{13}{12}$

Now, LHS = tan²B – sin² B

$=\left(\frac{5}{12}\right)^{2}-\left(\frac{5}{13}\right)^{2}$

$=\frac{25}{144}-\frac{25}{169}$

$=\frac{4225-3600}{144 \times 169}$

$=\frac{625}{144 \times 169}$

$=\frac{625}{24336}$

Now, RHS = sin⁴ B sec² B

$=\left(\frac{5}{13}\right)^{4}\left(\frac{13}{12}\right)^{2}$

$=\left(\frac{5}{13}\right)^{2}\left(\frac{5}{13}\right)^{2}\left(\frac{13}{12}\right)^{2}$

$=\frac{625}{144 \times 169}$

$=\frac{625}{24336}$

Now, LHS = RHS
Hence Proved

---

Question 44 

If $\cos \theta=\frac{\mathrm{q}}{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}}$, prove that 

$\left(\frac{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}}{\mathrm{p}}+\frac{\mathrm{q}}{\mathrm{p}}\right)^{2}= \frac{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}+\mathrm{q}}{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}-\mathrm{q}}}$

 

Sol :
Given: $\cos \theta=\frac{q}{\sqrt{p^{2}+q^{2}}}$
Now, squaring both the sides, we get

$=\cos ^{2} \theta=\frac{\mathrm{q}^{2}}{\mathrm{p}^{2}+\mathrm{q}^{2}}$

$\Rightarrow \mathrm{p}^{2}+\mathrm{q}^{2}=\frac{\mathrm{q}^{2}}{\cos ^{2} \theta}$

$\Rightarrow \mathrm{p}^{2}=\frac{\mathrm{q}^{2}}{\cos ^{2} \theta}-\mathrm{q}^{2}$

$\Rightarrow \mathrm{p}^{2}=\frac{\mathrm{q}^{2}-\mathrm{q}^{2} \cos ^{2} \theta}{\cos ^{2} \theta}$

$\Rightarrow \mathrm{p}^{2}=\frac{\mathrm{q}^{2}\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}$

$\Rightarrow \mathrm{p}^{2}=\frac{\mathrm{q}^{2} \sin ^{2} \theta}{\cos ^{2} \theta}$

⇒ p² = q² tan²θ …(1)

Now, solving LHS $=\left(\frac{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}}{\mathrm{p}}+\frac{\mathrm{q}}{\mathrm{p}}\right)^{2}$
Putting the value of p² in the above equation, we get

$=\left(\frac{\sqrt{q^{2} \tan ^{2} \theta+q^{2}}}{p}+\frac{q}{p}\right)^{2}$

$\Rightarrow\left(\frac{\sqrt{q^{2}\left(\tan ^{2} \theta+1\right)}}{p}+\frac{q}{p}\right)^{2}$

$\Rightarrow\left(\frac{\sqrt{\mathrm{q}^{2} \sec ^{2} \theta}}{\mathrm{p}}+\frac{\mathrm{q}}{\mathrm{p}}\right)^{2}$ [∵ 1+ tan² θ = sec² θ]

$\Rightarrow\left(\frac{\mathrm{q} \sec \theta}{\mathrm{p}}+\frac{\mathrm{q}}{\mathrm{p}}\right)^{2}$

$\Rightarrow \frac{\mathrm{q}^{2}(\sec \theta+1)^{2}}{\mathrm{p}^{2}}$

$\Rightarrow \frac{\mathrm{q}^{2}(\sec \theta+1)^{2}}{\mathrm{q}^{2} \tan ^{2} \theta}$ (from Eq. (1))

$\Rightarrow \frac{(\sec \theta+1)^{2}}{\left(\sec ^{2} \theta-1\right)}$

$\Rightarrow \frac{(\sec \theta+1)(\sec \theta+1)}{(\sec \theta-1)(\sec \theta+1)}$ [∵(a + b) (a – b) = (a² – b²)]

$\Rightarrow \frac{\sec \theta+1}{\sec \theta-1}$

Now, we solve the RHS

$=\frac{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}+\mathrm{q}}{\sqrt{\mathrm{p}^{2}+\mathrm{q}^{2}}-\mathrm{q}}$

$=\frac{\sqrt{\mathrm{q}^{2} \tan ^{2} \theta+\mathrm{q}^{2}}+\mathrm{q}}{\sqrt{\mathrm{q}^{2} \tan ^{2} \theta+\mathrm{q}^{2}}-\mathrm{q}}$

$=\frac{\sqrt{\mathrm{q}^{2}\left(\tan ^{2} \theta+1\right)}+\mathrm{q}}{\sqrt{\mathrm{q}^{2}\left(\tan ^{2} \theta+1\right)}-\mathrm{q}}$

$=\frac{\sqrt{\mathrm{q}^{2} \sec ^{2}}+\mathrm{q}}{\sqrt{\mathrm{q}^{2} \sec ^{2} \theta}-\mathrm{q}}$ [∵ 1+ tan² θ = sec² θ]

$=\frac{\operatorname{qsec} \theta+q}{\operatorname{qsec} \theta-q}$

$\Rightarrow \frac{\sec \theta+1}{\sec \theta-1}$

∴ LHS = RHS
Hence Proved

---

Question 45 

In the given figure, BC = 15 cm and sin B = 4/5, show that 

$\tan ^{2} \mathrm{B}-\frac{1}{\cos ^{2} \mathrm{B}}=-1$

Sol :
Given: BC =15cm and $\sin B=\frac{4}{5}$
We know that,
$\sin \theta=\frac{\text { side opposite to angle } \theta}{\text { hypotenuse }}$
Or $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$
$\sin \mathrm{B}=\frac{4}{5} \Rightarrow \frac{\mathrm{P}}{\mathrm{H}}=\frac{4}{5} \Rightarrow \frac{\mathrm{AC}}{\mathrm{AB}}=\frac{4}{5}$
Let,
Side opposite to angle B = 4k
and Hypotenuse = 5k
where, k is any positive integer
So, by Pythagoras theorem, we can find the third side of a triangle
⇒ (AC)² + (BC)² = (AB)²
⇒ (4k)² + (BC)² = (5)²
⇒ 16k² + (BC)² = 25k²
⇒ (BC)² = 25 k² –16 k²
⇒ (BC)² = 9 k²
⇒ BC =√9 k²
⇒ BC =±3k
But side BC can’t be negative. So, BC = 3k
Now, we have to find the value of cos B and tan B

We know that,
$\cos \theta=\frac{\text { side adjacent to angle } \theta}{\text { hypotenuse }}$
Side adjacent to angle B = BC =3k
Hypotenuse = AB =5k
So, $\cos B=\frac{3 k}{5 k}=\frac{3}{5}$
Now, tan B
We know that,
$\tan \theta=\frac{\text { side opposite to angle } \theta}{\text { side adjacent to angle } \theta}$
side opposite to angle B = AC =4k
Side adjacent to angle B = BC =3k
So, $\tan \mathrm{B}=\frac{4 \mathrm{k}}{3 \mathrm{k}}=\frac{4}{3}$

Now, $\tan ^{2} \mathrm{B}-\frac{1}{\cos ^{2} \mathrm{B}}$

$=\left(\frac{4}{3}\right)^{2}-\left(\frac{1}{\frac{3}{5}}\right)^{2}$

$=\frac{16}{9}-\frac{25}{9}$

$=\frac{-9}{9}$

= –1 = RHS
Hence Proved

---

Question 46 

In the given figure, find 3 tan θ – 2 sin α + 4 cos α.

Sol :
First of all, we find the value of RS
In right angled ∆RQS, we have
(RQ)² + (QS)² = (RS)²
⇒ (8)² + (6)² = (RS)²
⇒ 64 + 36 = (RS)²
⇒ RS =√100
⇒ RS =±10 [taking positive square root, since side cannot be negative]
⇒ RS =10

$\therefore \sin \alpha=\frac{P}{H}=\frac{8}{10}=\frac{4}{5}$

$\cos \alpha=\frac{B}{H}=\frac{6}{10}=\frac{3}{5}$

Now, we find the value of QP
In right angled ∆RQP
(RQ)² + (QP)² = (RP)²
⇒ (8)² + (QP)² = (17)²
⇒ 64 + (QP)² = 289
⇒ (QP)² =289–64
⇒ (QP)² =225
⇒ QP =√225
⇒ QP =±15 [taking positive square root, since side cannot be negative]
⇒ QP =15
tan θ$=\frac{\mathrm{P}}{\mathrm{B}}=\frac{8}{15}$

Now, 3 tan θ – 2 sinα + 4cos α
⇒ $3\left(\frac{8}{15}\right)-2\left(\frac{4}{5}\right)+4\left(\frac{3}{5}\right)$
⇒ $\frac{24}{15}-\frac{8}{5}+\frac{12}{5}$
⇒ $\frac{24-24+36}{15}$

⇒$\frac{36}{15}$

⇒$\frac{12}{5}$

---

Question 47 

In the given figure ΔABC is right angled at B and BD is perpendicular to AC. Find (i) cos θ, (ii) cot α.

Sol :

Firstly, we find the value of AC
In right angled ∆ABC
(AB)² + (BC)² = (AC)²
⇒ (12)² + (5)² = (AC)²
⇒ 144+25 =(AC)²
⇒ (AC)² =169
⇒ AC =√169
⇒ AC =±13
⇒ AC =13 [taking positive square root since, side cannot be negative]
(i) $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{12}{13}$
(ii) $\cot \alpha=\frac{\text { Base }}{\text { Perpendicular }}=\frac{12}{5}$

 

---

Question  48 

If 5 sin² θ + cos² θ = 2, find the value of sin θ.

 

Sol :
Given: 5 sin² θ + cos² θ = 2
⇒ 5 sin² θ + (1– sin² θ) = 2 [∵ sin² θ + cos² θ = 1]
⇒ 5 sin² θ + 1 – sin² θ = 2
⇒ 4 sin² θ = 2 – 1
⇒ 4 sin² θ = 1

$\Rightarrow \quad \sin ^{2} \theta=\frac{1}{4}$

$\Rightarrow \quad \sin \theta=\sqrt{\frac{1}{4}}$

$\Rightarrow \quad \sin \theta=\pm \frac{1}{2}$

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Question 49 

If 7 sin² θ + 3 cos² θ =4, find the value of tan θ.

 

Sol :
Given : 7 sin² θ + 3 cos² θ =4
⇒ 7 sin² θ + 3(1– sin² θ) = 4 [∵ sin² θ + cos² θ = 1]
⇒ 7 sin² θ + 3 –3 sin² θ = 4
⇒ 4 sin² θ = 4 – 3
⇒ 4 sin² θ = 1

$\Rightarrow \sin ^{2} \theta=\frac{1}{4}$

$\Rightarrow \sin \theta=\sqrt{\frac{1}{4}}$

$\Rightarrow \sin \theta=\pm \frac{1}{2}$

Put the value of $\sin ^{2} \theta=\frac{1}{4}$ in given equation, we get

$\Rightarrow 7\left(\frac{1}{2}\right)^{2}+3 \cos ^{2} \theta=4$

$\Rightarrow \frac{7}{4}+3 \cos ^{2} \theta=4$

$\Rightarrow 3 \cos ^{2} \theta=4-\frac{7}{4}$

$\Rightarrow 3 \cos ^{2} \theta=\frac{16-7}{4}$

$\Rightarrow 3 \cos ^{2} \theta=\frac{9}{4}$

$\Rightarrow \cos ^{2} \theta=\frac{3}{4}$

$\Rightarrow \cos \theta=\sqrt{\frac{3}{4}}$

$\Rightarrow \cos \theta=\pm \frac{\sqrt{3}}{2}$

Now, we know that tan θ $=\frac{\sin \theta}{\cos \theta}$

$\Rightarrow \tan \theta=\frac{\left(\pm \frac{1}{2}\right)}{\left(\pm \frac{\sqrt{3}}{2}\right)}$

$\Rightarrow \tan \theta=\pm \frac{1}{\sqrt{3}}$

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Question 50 

If 4 cos θ + 3 sin θ = 5, find the value of tan θ.

 

Sol :

Given : 4 cos θ+ 3 sin θ = 5
Squaring both the sides, we get
⇒ (4 cos θ+ 3 sin θ)² = 25
⇒ 16 cos² θ + 9 sin² θ + 2(4cos θ)(3sin θ)= 25 [∵ (a + b)² =a² +b² +2ab]
⇒ 16 cos² θ + 9 sin² θ + 24 cosθ sinθ = 25

Divide by cos² θ, we get
$\Rightarrow \frac{16 \cos ^{2} \theta}{\cos ^{2} \theta}+\frac{9 \sin ^{2} \theta}{\cos ^{2} \theta}+\frac{24 \cos \theta \sin \theta}{\cos ^{2} \theta}=\frac{25}{\cos ^{2} \theta}$
⇒ 16 + 9tan² θ + 24 tanθ = 25sec² θ
⇒ 16 + 9tan² θ + 24 tanθ = 25(1 + tan² θ) [∵ 1+ tan²θ = sec² θ]
⇒ 16 + 9tan² θ + 24 tanθ = 25+ 25 tan² θ
⇒ 16tan² θ – 24tanθ + 9 = 0
⇒ 16tan² θ – 12 tanθ – 12 tanθ +9 = 0
⇒ 4tanθ (4tan θ – 3) – 3(4tan θ – 3) = 0
⇒ (4tan θ – 3)² = 0
$\Rightarrow \tan \theta=\frac{3}{4}$

 

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Question 51 

If 7 sin A + 24 cos A = 25, find the value of tan A.

 

Sol :
Given : 7 sin A + 24 cos A = 25
Squaring both the sides, we get
⇒ (7 sin A + 24 cos A)² = 625
⇒ 49 sin² A +576 cos² A + 2(7sin A) (24cos A) = 625 [∵ (a + b)² =a² +b² +2ab]
⇒ 49 sin² A +576 cos² A + 336 cosA sinA = 625

Divide by cos² θ, we get
$\Rightarrow \frac{49 \sin ^{2} \mathrm{A}}{\cos ^{2} \mathrm{A}}+\frac{576 \cos ^{2} \mathrm{A}}{\cos ^{2} \mathrm{A}}+\frac{336 \cos \mathrm{A} \sin \mathrm{A}}{\cos ^{2} \mathrm{A}}=\frac{625}{\cos ^{2} \mathrm{A}}$

⇒ 49tan² A +576+ 336 tanA = 625sec² A
⇒ 49tan² A +576+ 336 tanA = 625(1 + tan² A) [∵ 1+ tan²θ = sec² θ]
⇒ 49tan² A +576+ 336 tanθA = 625+625 tan² A
⇒ 576tan² A – 336tanA + 49 = 0
⇒ 576tan² A – 168 tanA – 168 tanA +49 = 0
⇒ 24tanθ (24tan A – 7) – 7(24tan A – 7) = 0
⇒ (24tan A – 7)² = 0
$\Rightarrow \tan \mathrm{A}=\frac{7}{24}$

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Question 52 

If 9 sin θ + 40 cos θ= 41, find the value of cos θ and cosec θ

 

Sol :
Given: 9 sin θ + 40 cos θ= 41
⇒ 9sinθ = 41 – 40 cosθ …(i)
Squaring both sides, we get
⇒ 81sin² θ = 1681+1600 cos² θ – 2(41) (40cos θ) [∵ (a – b)² =a² +b² –2ab]
⇒ 81 (1– cos² θ) =1681+1600 cos² θ – 3280cosθ
⇒ 81 – 81cos² θ = 1681 +1600cos² θ – 3280 cosθ
⇒ 1681cos² θ –3280cos θ +1600 = 0
⇒ (41)² cos² θ – 2(41) (40cos θ) + (40)² = 0
⇒ (41cos θ – 40 )² = 0
$\Rightarrow \cos \theta=\frac{40}{41}$

Now, putting the value of cos θ in Eq. (i), we get

$\Rightarrow 9 \sin \theta=41-40\left(\frac{40}{41}\right)$

$\Rightarrow 9 \sin \theta=\left(\frac{1681-1600}{41}\right)$

$\Rightarrow \sin \theta=\left(\frac{81}{41 \times 9}\right)$

$\Rightarrow \frac{1}{\operatorname{cosec} \theta}=\left(\frac{9}{41}\right)$

$\Rightarrow \operatorname{cosec} \theta=\frac{41}{9}$

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Question 53 

If tan A + sec A = 3, find the value of sin A.

 

Sol :
tan A + sec A = 3
⇒ tanA = 3 – secA
Squaring both the sides, we get
⇒ tan² A =(3 – secA)²
⇒ tan² A = 9 + sec²A – 6sec A
⇒ sec² A – 1 = 9 + sec²A – 6sec A [∵ 1+ tan² A = sec² A]
⇒ –1 – 9 = –6secA
⇒ – 10 = –6sec A
$\Rightarrow \sec A=\frac{10}{6}$
$\Rightarrow \frac{1}{\cos A}=\frac{5}{3}\left[\because \sec A=\frac{1}{\cos A}\right]$
$\Rightarrow \cos A=\frac{3}{5}$

Now, tan A + sec A = 3
$\Rightarrow \frac{\sin A}{\cos A}+\frac{1}{\cos A}=3\left[\because \tan A=\frac{\sin A}{\cos A}\right]$
$\Rightarrow \frac{\sin A}{\cos A}=\frac{3 \cos A-1}{\cos A}$
⇒ sin A = 3cosA – 1

$\Rightarrow \sin A=3\left(\frac{3}{5}\right)-1$

$\Rightarrow \sin A=\left(\frac{9-5}{5}\right)$

$\Rightarrow \sin A=\left(\frac{4}{5}\right)$

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Question 54 

If cosec A + cot A = 5, find the value of cos A.

 

Sol :
cosec A + cot A = 5
⇒ cotA = 5 – cosecA
Squaring both the sides, we get
⇒ cot² A =(5 – cosecA)²
⇒ cot² A = 25 + cosec²A – 10cosec A
⇒ cosec² A – 1 = 25 + cosec²A – 10cosec A [∵ 1+ cot² A = cosec² A]
⇒ –1 – 25 = –10cosecA
⇒ – 26 = –10cosec A
$\Rightarrow \operatorname{cosec} A=\frac{26}{10}$
$\Rightarrow \frac{1}{\sin A}=\frac{13}{5}\left[\because \operatorname{cosec} A=\frac{1}{\sin A}\right]$
$\Rightarrow \sin \mathrm{A}=\frac{5}{13}$

Now, cosec A + cot A = 5
$\Rightarrow \frac{1}{\sin A}+\frac{\cos A}{\sin A}=5\left[\because \cot A=\frac{\cos A}{\sin A}\right]$
$\Rightarrow \frac{13}{5}+\frac{\cos A}{\frac{5}{13}}=5$

$\Rightarrow \frac{13}{5}+\frac{13 \cos A}{5}=5$

$\Rightarrow \frac{13 \cos A}{5}=5-\frac{13}{5}$

$\Rightarrow \frac{13 \cos A}{5}=\frac{25-13}{5}$

$\Rightarrow \cos A=\frac{12}{13}$

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Question 55 

If tan θ + sec θ = x, show that 

$\sin \theta=\frac{x^{2}-1}{x^{2}+1}$

Sol :
tan θ+ sec θ = x
⇒ tan θ = x – sec θ
Squaring both sides, we get
⇒ tan² θ =(x – secθ)²
⇒ tan² θ = x² + sec²θ – 2xsec θ
⇒ sec² θ – 1 = x² + sec²θ – 2xsec θ [∵ 1+ tan² A = sec² A]
⇒ –1 – x² = –2xsecθ
$\Rightarrow \sec \theta=\frac{1+x^{2}}{2 x}$
Now,
tan θ = x – sec θ
$\Rightarrow \frac{\sin \theta}{\cos \theta}=x-\sec \theta$
$\Rightarrow \sin \theta\left(\frac{1+x^{2}}{2 x}\right)=x-\left(\frac{1+x^{2}}{2 x}\right)$
$\Rightarrow \sin \theta\left(\frac{1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)=\left(\frac{2 \mathrm{x}^{2}-1+\mathrm{x}^{2}}{2 \mathrm{x}}\right)$
$\Rightarrow \sin \theta\left(\frac{1+x^{2}}{2 x}\right)=\left(\frac{x^{2}-1}{2 x}\right)$
$\Rightarrow \sin \theta=\frac{x^{2}-1}{x^{2}+1}$ = RHS
Hence Proved

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Question 56 

If cos θ +sin θ=1, prove that cos θ – sin θ = ± 1

 

Sol :

Using the formula,
(a+b)² + (a – b)² = 2(a²+b²)
⇒ (cos θ +sin θ)² + (cos θ – sin θ)² = 2(cos²θ + sin² θ)
⇒ 1 + (cos θ – sin θ)² = 2(1)
⇒ (cos θ – sin θ)² = 2 –1
⇒ (cos θ – sin θ)² = 1
⇒ (cos θ – sin θ) =√1
⇒ (cos θ – sin θ) = ±1

 

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S.no	Chapters	Links
1	Real numbers	Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4
2	Polynomials	Exercise 2.1 Exercise 2.2 Exercise 2.3
3	Pairs of Linear Equations in Two Variables	Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5
4	Trigonometric Ratios and Identities	Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4
5	Triangles	Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5
6	Statistics	Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4
7	Quadratic Equations	Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5
8	Arithmetic Progressions (AP)	Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4
9	Some Applications of Trigonometry: Height and Distances	Exercise 9.1
10	Coordinates Geometry	Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4
11	Circles	Exercise 11.1 Exercise 11.2
12	Constructions	Exercise 12.1
13	Area related to Circles	Exercise 13.1
14	Surface Area and Volumes	Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4
15	Probability	Exercise 15.1