KC Sinha Mathematics Solution Class 10 Chapter 4 Trigonometric Ratios and Identities Exercise 4.4

Exercise 4.1
Exercise 4.2
Exercise 4.3
Exercise 4.4

Exercise 4.4

Question 1

Fill in the blanks

(i) sin² θ cosec² θ = ……..
(ii) 1 + tan² θ = ……
(iii) Reciprocal sin θ. cot θ = ……
(iv) 1–.......= cos²θ
(v) $\tan A=\frac{\cdots \cdots \cdot}{\cos A}$
(vi) $\ldots \ldots=\frac{\cos A}{\sin A}$
(vii) cos θ is reciprocal of .........
(viii) Reciprocal of sin θ is.........
(ix) Value of sin θ in terms of cos θ is
(x) Value of cos θ in terms of sin θ is
Sol :
(i) Given: sin² θ cosec² θ

$\Rightarrow \sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta}$ $\left[\because \sin \theta=\frac{1}{\csc \theta}\right]$
= 1

(ii) Given: 1 + tan² θ

$=1+\left(\frac{\sin \theta}{\cos \theta}\right)^{2}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=1+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos ^{2} \theta}$[∵ cos² θ + sin² θ = 1]
$=\frac{1}{\cos ^{2} \theta}$
= sec² θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$

(iii) Given : sin θ cot θ
Firstly, we simplify the given trigonometry

$=\sin \theta \times \frac{\cos \theta}{\sin \theta}$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
= cos θ
Now, the reciprocal of cos θ is
$=\frac{1}{\cos \theta}$
=sec θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$

(iv) Given: 1 – x = cos²θ
Subtracting 1 to both the sides, we get
1 – x –1 = cos²θ – 1
⇒ –x = – sin²θ
⇒ x =sin² θ

(v) $\tan A=\frac{\sin A}{\cos A}$

(vi) $\cot A=\frac{\cos A}{\sin A}$

(vii) $\cos \theta=\frac{1}{\sec \theta}$

(viii) $\sin \theta=\frac{1}{\cos \theta}$

(ix) We know that
cos² θ + sin² θ = 1
⇒ sin² θ = 1 – cos² θ
⇒ sin θ = √(1 – cos² θ)
(x) We know that
cos² θ + sin² θ = 1
⇒ cos ² θ = 1 – sin² θ
⇒ cos θ = √(1 – sin² θ)

Question 2

If sin θ= p and cos θ = q, what is the relation between p and q ?

Sol :
We know that,
cos² θ + sin² θ = 1 …(i)
Given : sin θ = p and cos θ = q
Putting the values of sin θ and cos θ in eq. (i), we get
(q)² + (p)² =1
⇒ p² + q² =1

Question 3

If cos A = x, express sin A in terms of x

Sol :
We know that
cos² θ + sin² θ = 1
⇒ sin² θ = 1 – cos² θ
⇒ sin θ = √(1 – cos² θ)
And Given that cos θ = x
⇒ sin θ = √(1– x²)

Question 4

If x cos θ = 1 and y sin θ = 1 find the value of tan θ.

Sol :
Given x cosθ = 1 and y sinθ = 1

$\Rightarrow \cos \theta=\frac{1}{x}$ and $\sin \theta=\frac{1}{\mathrm{y}}$

Now, we know that

$\tan \theta=\frac{\sin \theta}{\cos \theta}$

Putting the value of sin θ and cos θ, we get
$\tan \theta=\frac{\frac{1}{y}}{\frac{1}{x}}$
$\Rightarrow \tan \theta=\frac{x}{y}$

Question 5

If cos40^o = p, then write the value of sin 40^o in terms of p.

Sol :
We know that
cos² θ + sin² θ = 1
⇒ cos² 40° + sin² 40° = 1
⇒ sin² 40° = 1 – cos² 40°
⇒ sin 40° = √(1 – cos² 40°)
And Given that cos 40° = p
⇒ sin 40° = √(1– p²)

Question 6

If sin 77° = x, then write the value of cos 77^o in terms of x.

Sol :
We know that
cos² θ + sin² θ = 1
⇒ cos² 77° + sin² 77° = 1
⇒ cos ² 77° = 1 – sin² 77°
⇒ cos 77° = √(1 – sin² 77°)
And Given that sin 77° = x
⇒ cos 77° = √(1 – x² )

Question 7

If cos55° = x², then write the value of sin 55^o in terms of x.

Sol :
We know that
cos² θ + sin² θ = 1
⇒ cos² 55° + sin² 55° = 1
⇒ sin² 55° = 1 – cos² 55°
⇒ sin 55° = √(1 – cos² 55°)
And Given that cos 55° = x²
⇒ sin 55° = √{1– (x²)²}
⇒ sin 55° = √(1 – x⁴)

Question 8

If, sin 50° = α then write the value of cos 50° in terms of α.

Sol :
We know that
cos² θ + sin² θ = 1
⇒ cos² 50° + sin² 50° = 1
⇒ cos ² 50° = 1 – sin² 50°
⇒ cos 50° = √(1 – sin² 50°)
And Given that sin 50° = a
⇒ cos 50° = √(1 – a² )

Question 9

If x cos A = 1 and tan A = y, then what is the value of x² – y².

Sol :
Given x cos A = 1 and tan A =y
$\Rightarrow \mathrm{x}=\frac{1}{\cos \mathrm{A}}$ and $\frac{\sin A}{\cos A}=y$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
To find: x² – y²

Putting the values of x and y , we get
$\left(\frac{1}{\cos A}\right)^{2}-\left(\frac{\sin A}{\cos A}\right)^{2}$
$=\frac{1}{\cos ^{2} A}-\frac{\sin ^{2} A}{\cos ^{2} A}$
$=\frac{1-\sin ^{2} A}{\cos ^{2} A}$
$=\frac{\cos ^{2} A}{\cos ^{2} A}$ [∵ cos² θ + sin² θ = 1]
= 1

Question 10

Prove the followings identities:

(1 – sin θ)(1 + sin θ) = cso² θ
Sol :
Taking LHS = (1 – sinθ)(1+ sinθ)
Using identity , (a + b) (a – b) = (a² – b²) , we get
= (1)² – (sinθ)²
= 1 – sin² θ
= cos² θ [∵ cos² θ + sin² θ = 1]
= RHS
Hence Proved

Question 11

Prove the followings identities:

(1 + cos θ)(1 – cos θ) = sin²θ
Sol :
Taking LHS =(1 – cosθ)(1+ cosθ)
Using identity , (a + b) (a – b) = (a² – b²) , we get
= (1)² – (cosθ)²
= 1 – cos² θ
= sin² θ [∵ cos² θ + sin² θ = 1]
= RHS
Hence Proved

Question 12

Prove the followings identities:

$\frac{(1-\cos \theta)(1+\cos \theta)}{(1-\sin \theta)(1+\sin \theta)}=\tan ^{2} \theta$

Sol :
Taking LHS $\frac{(1-\cos \theta)(1+\cos \theta)}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{(1)^{2}-(\cos \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}$ [Using identity , (a + b) (a – b) = (a² – b²)]
$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
= tan² θ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
= RHS
Hence Proved

Question 13

Prove the followings identities:

$\frac{1}{\sec \theta+\tan \theta}=\sec \theta-\tan \theta$

Sol :
Taking LHS $=\frac{1}{\sec \theta+\tan \theta}$
Multiplying and divide by the conjugate of sec θ + tan θ
$=\frac{1}{\sec \theta+\tan \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}$
$=\frac{\sec \theta-\tan \theta}{\sec ^{2} \theta-\tan ^{2} \theta}$ [Using identity , (a + b) (a – b) = (a² – b²)]
= sec θ – tan θ [∵ 1+ tan² θ = sec² θ ]
= RHS
Hence Proved