| Exercise
4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
Exercise 4.3
Question 1
Express the following as trigonometric ratio of complementary angle of θ.
(i) cos θ (ii) sec θ(iii) cot θ (iv) cosec θ
(v) tan θ
Sol :
(i) We know that
$\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}=\frac{A B}{A C}=\sin \left(90^{\circ}-\theta\right)$
⇒ cosθ = sin (90° - θ)
$\cos \theta=\frac{\text { base }}{\text { Hypotenuse }}=\frac{A B}{A C}=\sin \left(90^{\circ}-\theta\right)$
⇒ cosθ = sin (90° - θ)
(ii) We know that
$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}=\frac{A C}{A B}=\operatorname{cosec}\left(90^{\circ}-\theta\right)$
$\sec \theta=\frac{\text { hypotenuse }}{\text { base }}=\frac{A C}{A B}=\operatorname{cosec}\left(90^{\circ}-\theta\right)$
(iii) We know that
$\cot \theta=\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\tan \left(90^{\circ}-\theta\right)$
$\cot \theta=\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\tan \left(90^{\circ}-\theta\right)$
(iv) We know that
$\operatorname{cosec} \theta=\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{B C}=\sec \left(90^{\circ}-\theta\right)$
(v) We know that
$\tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{\mathrm{BC}}{\mathrm{AB}}=\cot \left(90^{\circ}-\theta\right)$
Question 2
Express the following as trigonometric ratio of complementary angle of 90o- θ.
(i) tan (90o- θ)(ii) cos (90o- θ)
Sol :
(i) We know that,
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
$\Rightarrow \tan \left(90^{\circ}-\theta\right)=\frac{\sin \left(90^{\circ}-\theta\right)}{\cos \left(90^{\circ}-\theta\right)}$
$\Rightarrow \tan \left(90^{\circ}-\theta\right)=\frac{\sin 90^{\circ} \cos \theta-\cos 90^{\circ} \sin \theta}{\cos 90^{\circ} \cos \theta+\sin 90^{\circ} \sin \theta}$[∵ sin 90° = 1 and cos 90° = 0]
$\Rightarrow \tan \left(90^{\circ}-\theta\right)=\frac{\cos \theta}{\sin \theta}$
⇒ tan (90° - θ ) = cot θ
(ii) We know that.
Cos(A - B) = cos A cos B + sin A sin B
⇒ cos (90o- θ) = cos 90° cos θ + sin 90° sin θ
⇒ cos (90o- θ) = (0) cos θ + (1) sin θ
⇒ cos (90o- θ) = sin θ
(iii) cos 48o=sin (…) (iv) cos 70o = sin (…)
(v) cos 50o = sin (…) (vi) sec 32o=cosec(…)
Sol :
(i) We know that
Sin θ = cos (90° - θ)
Here, θ = 70°
⇒ sin 70° = cos(90° -70°)
⇒ sin 70° = cos 20°
Cos(A - B) = cos A cos B + sin A sin B
⇒ cos (90o- θ) = cos 90° cos θ + sin 90° sin θ
⇒ cos (90o- θ) = (0) cos θ + (1) sin θ
⇒ cos (90o- θ) = sin θ
Question 3
Fill up the blanks by an angle between 0oand 90o:
(i) sin 70o = cos(…) (ii) sin 35o = cos(…)(iii) cos 48o=sin (…) (iv) cos 70o = sin (…)
(v) cos 50o = sin (…) (vi) sec 32o=cosec(…)
Sol :
(i) We know that
Sin θ = cos (90° - θ)
Here, θ = 70°
⇒ sin 70° = cos(90° -70°)
⇒ sin 70° = cos 20°
(ii) We know that
Sin θ = cos (90° - θ)
Here, θ = 35°
⇒ sin 35° = cos(90° -35°)
⇒ sin 35° = cos 55°
Sin θ = cos (90° - θ)
Here, θ = 35°
⇒ sin 35° = cos(90° -35°)
⇒ sin 35° = cos 55°
(iii) cos θ = sin (90° - θ)
Here, θ = 48°
⇒ cos 48° = sin (90° - 48°)
⇒ cos 48° = sin 42°
Here, θ = 48°
⇒ cos 48° = sin (90° - 48°)
⇒ cos 48° = sin 42°
(iv) cos θ = sin (90° - θ)
Here, θ = 70°
⇒ cos 70° = sin (90° - 70°)
⇒ cos 70° = sin 20°
Here, θ = 70°
⇒ cos 70° = sin (90° - 70°)
⇒ cos 70° = sin 20°
(v) cos θ = sin (90° - θ)
Here, θ = 50°
⇒ cos 50° = sin (90° - 50°)
⇒ cos 50° = sin 40°
Here, θ = 50°
⇒ cos 50° = sin (90° - 50°)
⇒ cos 50° = sin 40°
(vi) sec θ = cosec (90°-θ)
Here, θ = 32°
⇒ sec 32° = cosec(90° – 32°)
⇒ sec 32° = cosec 58°
(iii) sec A =… (iv) tan B =…
(v) cosec B =… (vi) cot A=…
Sol :
(i) Here, A+B = 90°
⇒ A = 90° - B
Multiplying both sides by Sin, we get
Sin A = Sin (90° - B)
⇒ sin A = Cos B [∵ cos θ = sin (90° - θ)]
(ii) Here, A+B = 90°
⇒ B = 90° - A
Multiplying both sides by cos, we get
Cos B = cos (90° - A)
⇒ cos B = sin A [∵ Sin θ = cos (90° - θ)]
(iii) Here, A+B = 90°
⇒ A = 90° - B
Multiplying both sides by sec, we get
Sec A = Sec (90° - B)
⇒ sec A = Cosec B [∵ cosec θ = sec (90° - θ)]
(iv) Here, A+B = 90°
⇒ B = 90° - A
Multiplying both sides by tan, we get
tan B = tan (90° - A)
⇒ tan B = cot A [∵ cot θ = tan (90° - θ)]
(v) Here, A+B = 90°
⇒ B = 90° - A
Multiplying both sides by cosec, we get
Cosec B = cosec (90° - A)
⇒ cosec B = sec A [∵ sec θ = cosec (90° - θ)]
(vi) Here, A+B = 90°
⇒ A = 90° - B
Multiplying both sides by Sin, we get
cotA = cot (90° - B)
⇒ cot A = tan B [∵ tan θ = cot (90° - θ)]
Given sin 37° = a
We know that sin θ = cos (90° - θ)
Here, θ = 37°
⇒ cos (90° - 37°) = a
⇒ cos 53° = a
Given cos 47° = a
We know that cos θ = sin (90° - θ)
Here, θ = 47°
⇒ sin (90° - 47°) = a
⇒ sin 43° = a
Given sin 52° = a
We know that sin θ = cos (90° - θ)
Here, θ = 52°
⇒ cos (90° - 52°) = a
⇒ cos 38° = a
Sol :
Given sin 56° = x
We know that sin θ = cos (90° - θ)
Here, θ = 56°
⇒ cos (90° - 56°) = x
⇒ cos 34° = x
Here, θ = 32°
⇒ sec 32° = cosec(90° – 32°)
⇒ sec 32° = cosec 58°
Question 4
If A+B=90o, then fill up the blanks with suitable trigonometric ratio of complementary angle of A
or B.
(i) sin A =…. (ii) cos B =…(iii) sec A =… (iv) tan B =…
(v) cosec B =… (vi) cot A=…
Sol :
(i) Here, A+B = 90°
⇒ A = 90° - B
Multiplying both sides by Sin, we get
Sin A = Sin (90° - B)
⇒ sin A = Cos B [∵ cos θ = sin (90° - θ)]
(ii) Here, A+B = 90°
⇒ B = 90° - A
Multiplying both sides by cos, we get
Cos B = cos (90° - A)
⇒ cos B = sin A [∵ Sin θ = cos (90° - θ)]
(iii) Here, A+B = 90°
⇒ A = 90° - B
Multiplying both sides by sec, we get
Sec A = Sec (90° - B)
⇒ sec A = Cosec B [∵ cosec θ = sec (90° - θ)]
(iv) Here, A+B = 90°
⇒ B = 90° - A
Multiplying both sides by tan, we get
tan B = tan (90° - A)
⇒ tan B = cot A [∵ cot θ = tan (90° - θ)]
(v) Here, A+B = 90°
⇒ B = 90° - A
Multiplying both sides by cosec, we get
Cosec B = cosec (90° - A)
⇒ cosec B = sec A [∵ sec θ = cosec (90° - θ)]
(vi) Here, A+B = 90°
⇒ A = 90° - B
Multiplying both sides by Sin, we get
cotA = cot (90° - B)
⇒ cot A = tan B [∵ tan θ = cot (90° - θ)]
Question 5 A
If sin 37o=a, then express cos 53o in terms of a.
Sol :Given sin 37° = a
We know that sin θ = cos (90° - θ)
Here, θ = 37°
⇒ cos (90° - 37°) = a
⇒ cos 53° = a
Question 5 B
If cos 47o=a, then express sin 43o in terms of a.
Sol :Given cos 47° = a
We know that cos θ = sin (90° - θ)
Here, θ = 47°
⇒ sin (90° - 47°) = a
⇒ sin 43° = a
Question 5 C
If sin 52o=a, then express sin 38o in terms of a.
Sol :Given sin 52° = a
We know that sin θ = cos (90° - θ)
Here, θ = 52°
⇒ cos (90° - 52°) = a
⇒ cos 38° = a
Question 5 D
If sin 56o=x, then express sin 34o in terms of x.
Given sin 56° = x
We know that sin θ = cos (90° - θ)
Here, θ = 56°
⇒ cos (90° - 56°) = x
⇒ cos 34° = x
Question 6
Find the value of
(i) $\frac{\cos 59^{\circ}}{\sin 31^{\circ}}$
(ii) $\frac{\cos 53^{\circ}}{\sin 37^{\circ}}$
(iii) $\frac{\sin 20^{\circ}}{\cos
70^{\circ}}$
(iv) $\frac{\sqrt{2} \sin 22^{\circ}}{\cos
68^{\circ}}$
(v)$\frac{\sin 10^{\circ}}{\cos 80^{\circ}}$
(vi) $\frac{\sin 27^{\circ}}{\cos
63^{\circ}}$
(vii) $\frac{\sqrt{3} \cos 65^{\circ}}{\sin 25^{\circ}}$
(viii) $\frac{\cos 29^{\circ}}{\sin
61^{\circ}}$
(ix) sin 54° – cos 36°
(x) $\frac{\tan 80^{\circ}}{\cot
10^{\circ}}$
(xi) cosec 31° – sec 59°
(xi) cosec 31° – sec 59°
(xii) $\frac{\sin 18^{\circ}}{\cos
72^{\circ}}$
(xiii) $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$
Sol :
(i)$\frac{\cos 59^{\circ}}{\sin 31^{\circ}}=\frac{\sin \left(90^{\circ}-59^{\circ}\right)}{\sin 31^{\circ}}$
(xiii) $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$
Sol :
(i)$\frac{\cos 59^{\circ}}{\sin 31^{\circ}}=\frac{\sin \left(90^{\circ}-59^{\circ}\right)}{\sin 31^{\circ}}$
$=\frac{\sin 31^{\circ}}{\sin
31^{\circ}}=1$ [∵ cos θ = sin (90° - θ)]
(ii) $\frac{\cos 53^{\circ}}{\sin 37^{\circ}}=\frac{\sin \left(90^{\circ}-53^{\circ}\right)}{\sin 37^{\circ}}$
$=\frac{\sin 37^{\circ}}{\sin
37^{\circ}}=1$ [∵ cos θ = sin (90° - θ)]
(iii) $\frac{\sin 20^{\circ}}{\cos 70^{\circ}}=\frac{\cos \left(90^{\circ}-20^{\circ}\right)}{\cos 70^{\circ}}$
$=\frac{\cos 70^{\circ}}{\cos
70^{\circ}}=1$ [∵ Sin θ = cos (90° - θ)]
(iv) $\frac{\sqrt{2} \sin 22^{\circ}}{\cos 68^{\circ}}=\frac{\sqrt{2} \cos \left(90^{\circ}-22^{\circ}\right)}{\cos
68^{\circ}}$
$=\frac{\sqrt{2} \cos 68^{\circ}}{\cos
68^{\circ}}=\sqrt{2}$ [∵ Sin θ = cos (90° - θ)]
(v) $\frac{\sin 10^{\circ}}{\cos 80^{\circ}}=\frac{\cos \left(90^{\circ}-10^{\circ}\right)}{\cos 80^{\circ}}$
$=\frac{\cos 80^{\circ}}{\cos
80^{\circ}}=1$ [∵ Sin θ = cos (90° - θ)]
(vi) $\frac{\sin 27^{\circ}}{\cos 63^{\circ}}=\frac{\cos \left(90^{\circ}-27^{\circ}\right)}{\cos 63^{\circ}}$
$=\frac{\cos 63^{\circ}}{\cos
63^{\circ}}=1$ [∵ Sin θ = cos (90° - θ)]
(vii) $\frac{\sqrt{3} \cos 65^{\circ}}{\sin 25^{\circ}}=\frac{\sqrt{3} \sin \left(90^{\circ}-65^{\circ}\right)}{\sin
25^{\circ}}$
$=\frac{\sin 25^{\circ}}{\sin
25^{\circ}}=\sqrt{3}$ [∵ cos θ = sin (90° - θ)]
(viii) $\frac{\cos 29^{\circ}}{\sin 61^{\circ}}=\frac{\sin \left(90^{\circ}-29^{\circ}\right)}{\sin 61^{\circ}}$
$=\frac{\sin 61^{\circ}}{\sin
61^{\circ}}=1$ [∵ cos θ = sin (90° - θ)]
(ix) sin 54° - sin(90° - 36°) [∵ cos θ = sin (90° - θ)]
⇒ sin 54° - sin 54°
⇒ 0
⇒ sin 54° - sin 54°
⇒ 0
(x) $\frac{\tan 80^{\circ}}{\cot 10^{\circ}}=\frac{\cot \left(90^{\circ}-80^{\circ}\right)}{\cot 10^{\circ}}$
$=\frac{\cot 10^{\circ}}{\cot
10^{\circ}}=1$ [∵ tan θ = cot (90° - θ)]
(xi) cosec 31° - cosec(90° - 59°) [∵ sec θ = cosec (90° - θ)]
⇒ cosec 31° - cosec 31°
⇒ 0
⇒ cosec 31° - cosec 31°
⇒ 0
(xii) $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}=\frac{\cos \left(90^{\circ}-18^{\circ}\right)}{\cos 72^{\circ}}$
$=\frac{\cos 72^{\circ}}{\cos
72^{\circ}}=1$ [∵ Sin θ = cos (90° - θ)]
(xiii) $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}=\frac{\cot \left(90^{\circ}-65^{\circ}\right)}{\cot 25^{\circ}}$
$=\frac{\cot 25^{\circ}}{\cot
25^{\circ}}=1$ [∵ tan θ = cot (90° - θ)]
(ii) If cos 44o = 0.7193, then sin 46o=…..
(iii) sin 50o+cos 40o = 2 sin (………)
(iv) Value of $\frac{\sin 70^{\circ}}{\cos 20^{\circ}}$ is ………
Sol :
(i) Given: sin 50o=0.7660
We know that
Sin θ = cos (90° - θ)
⇒ cos (90° - 50°) = 0.7660
⇒ cos 40° = 0.7660
Question 7
Fill up the blanks :
(i) If sin 50o=0.7660, then cos 40o=……(ii) If cos 44o = 0.7193, then sin 46o=…..
(iii) sin 50o+cos 40o = 2 sin (………)
(iv) Value of $\frac{\sin 70^{\circ}}{\cos 20^{\circ}}$ is ………
Sol :
(i) Given: sin 50o=0.7660
We know that
Sin θ = cos (90° - θ)
⇒ cos (90° - 50°) = 0.7660
⇒ cos 40° = 0.7660
(ii) Given: cos 44° = 0.7193
We know that,
cos θ = sin (90° - θ)
⇒ sin (90° - 44°) = 0.7193
⇒ sin 46° = 0.7193
We know that,
cos θ = sin (90° - θ)
⇒ sin (90° - 44°) = 0.7193
⇒ sin 46° = 0.7193
(iii) LHS = sin 50° + cos 40°
⇒ sin 50° + sin (90° - 40°) [∵ cos θ = sin (90° - θ)]
⇒ sin 50° + sin 50°
⇒ 2sin 50°
⇒ sin 50° + sin (90° - 40°) [∵ cos θ = sin (90° - θ)]
⇒ sin 50° + sin 50°
⇒ 2sin 50°
(iii) $\frac{\sin 70^{\circ}}{\cos 20^{\circ}}=\frac{\cos \left(90^{\circ}-70^{\circ}\right)}{\cos 20^{\circ}}$
$=\frac{\cos 20^{\circ}}{\cos
20^{\circ}}=1$ [∵ Sin θ = cos (90° - θ)]
Given: A+B =90°
⇒ B = 90° - A
Multiplying both side by cos, we get
= cos B = Cos (90° - A)
⇒ cos B = sin A [∵ Sin θ = cos (90° - θ)]
Given: X+Y =90°
⇒ X= 90° - Y
Multiplying both side by cos, we get
= cos X = Cos (90° - Y)
⇒ cos X = sin Y [∵ Sin θ = cos (90° - θ)]
(b) $\tan A=\frac{a}{b}$
Sol :
(a) LHS = a2 +b2
= (sin A)2 + (sin B)2
= sin2 A + sin2 B
= sin2 A + sin2 (90° - A) [∵ cos θ = sin (90° - θ)]
= sin2 A + cos2 A
= 1 [∵ sin2 θ + cos2 θ = 1]
=RHS
Hence Proved
Question 8 A
If A + B = 90o, then express cos B in terms of simplest trigonometric ratio of A.
Sol :Given: A+B =90°
⇒ B = 90° - A
Multiplying both side by cos, we get
= cos B = Cos (90° - A)
⇒ cos B = sin A [∵ Sin θ = cos (90° - θ)]
Question 8 B
If X + Y = 90o, then express cos X in terms of simplest trigonometric ratio of Y.
Sol :Given: X+Y =90°
⇒ X= 90° - Y
Multiplying both side by cos, we get
= cos X = Cos (90° - Y)
⇒ cos X = sin Y [∵ Sin θ = cos (90° - θ)]
Question 9 A
If A + B = 90o, sin A = a, sin B = b, then prove that
(a) a2 + b2 = 1(b) $\tan A=\frac{a}{b}$
Sol :
(a) LHS = a2 +b2
= (sin A)2 + (sin B)2
= sin2 A + sin2 B
= sin2 A + sin2 (90° - A) [∵ cos θ = sin (90° - θ)]
= sin2 A + cos2 A
= 1 [∵ sin2 θ + cos2 θ = 1]
=RHS
Hence Proved
(b) LHS = tan A
Now, taking RHS $=\frac{a}{b}$
⇒ $\Rightarrow \frac{\sin A}{\sin B}$
$\Rightarrow \frac{\sin A}{\sin \left(90^{\circ}-A\right)}$ {given, A +B = 90°)
$\Rightarrow \frac{\sin A}{\cos A}$ [∵ cos θ = sin (90° - θ)]
⇒ tan A
=LHS
∴ LHS = RHS
Hence Proved
LHS = sin (50° + θ) – cos (40° - θ)
We know that,
Sin A = cos (90° - A)
Here, A = 50° + θ
⇒ cos {90° -( 50° + θ)} – cos (40° - θ)
⇒ cos (40° - θ) – cos (40° - θ)
= 0 = RHS
Hence Proved
Taking LHS,
$\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)}$ [∵ cos θ = sin (90° - θ) and Sin θ = cos (90° - θ)]
$\Rightarrow \frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\sin \theta}$
⇒ 1+ 1
= 2 = RHS
Hence Proved
Sol :
Now, taking RHS $=\frac{a}{b}$
⇒ $\Rightarrow \frac{\sin A}{\sin B}$
$\Rightarrow \frac{\sin A}{\sin \left(90^{\circ}-A\right)}$ {given, A +B = 90°)
$\Rightarrow \frac{\sin A}{\cos A}$ [∵ cos θ = sin (90° - θ)]
⇒ tan A
=LHS
∴ LHS = RHS
Hence Proved
Question 9 B
Show that sin(50° + θ) – cos (40° – θ) = 0.
Sol :LHS = sin (50° + θ) – cos (40° - θ)
We know that,
Sin A = cos (90° - A)
Here, A = 50° + θ
⇒ cos {90° -( 50° + θ)} – cos (40° - θ)
⇒ cos (40° - θ) – cos (40° - θ)
= 0 = RHS
Hence Proved
Question 10
Prove that $\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\sin \theta}{\cos
\left(90^{\circ}-\theta\right)}=2$
Sol :Taking LHS,
$\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)}$ [∵ cos θ = sin (90° - θ) and Sin θ = cos (90° - θ)]
$\Rightarrow \frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\sin \theta}$
⇒ 1+ 1
= 2 = RHS
Hence Proved
Question 11 A
In a ∆ABC prove that
$\sin \frac{\mathrm{B}+\mathrm{C}}{2}=\cos \frac{\mathrm{A}}{2}$Sol :
In ∆ABC,
Sum of angles of a triangle = 180°
A + B + C = 180°
⇒ B + C = 180° - A
Sum of angles of a triangle = 180°
A + B + C = 180°
⇒ B + C = 180° - A
Multiplying both sides
by $\frac{1}{2}$
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}-\mathrm{A}}{2}$
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}}{2}-\frac{\mathrm{A}}{2}$
$=\frac{B+C}{2}=90^{\circ}-\frac{A}{2}$ …(1)
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}-\mathrm{A}}{2}$
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}}{2}-\frac{\mathrm{A}}{2}$
$=\frac{B+C}{2}=90^{\circ}-\frac{A}{2}$ …(1)
Taking LHS
$\sin \frac{\mathrm{B}+\mathrm{C}}{2}$
$=\sin \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)$ (from eq (1))
$=\cos \frac{A}{2}$ [∵ sin (90° - θ) = cos θ]
=RHS
Hence Proved
Sol :
$\sin \frac{\mathrm{B}+\mathrm{C}}{2}$
$=\sin \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)$ (from eq (1))
$=\cos \frac{A}{2}$ [∵ sin (90° - θ) = cos θ]
=RHS
Hence Proved
Question 11 B
In a ∆ABC prove that
$\tan \frac{\mathrm{B}+\mathrm{C}}{2}=\cot \frac{\mathrm{A}}{2}$Sol :
In ∆ABC,
Sum of angles of a triangle = 180°
A + B + C = 180°
⇒ B + C = 180° - A
Sum of angles of a triangle = 180°
A + B + C = 180°
⇒ B + C = 180° - A
Multiplying both sides by
$\frac{1}{2}$
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}-\mathrm{A}}{2}$
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}}{2}-\frac{\mathrm{A}}{2}$
$=\frac{B+C}{2}=90^{\circ}-\frac{A}{2}$ …(2)
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}-\mathrm{A}}{2}$
$=\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}}{2}-\frac{\mathrm{A}}{2}$
$=\frac{B+C}{2}=90^{\circ}-\frac{A}{2}$ …(2)
Taking LHS
$\tan \frac{\mathrm{B}+\mathrm{C}}{2}$
$=\tan \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)$ (from eq (2))
$=\cot \frac{A}{2}$ [∵ tan (90° - θ) = cot θ]
=RHS
Hence Proved
Sol :
$\tan \frac{\mathrm{B}+\mathrm{C}}{2}$
$=\tan \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)$ (from eq (2))
$=\cot \frac{A}{2}$ [∵ tan (90° - θ) = cot θ]
=RHS
Hence Proved
Question 11 C
In a ∆ABC prove that
$\cos \frac{A+B}{2}=\sin \frac{C}{2}$Sol :
In ∆ABC,
Sum of angles of a triangle = 180°
A + B + C = 180°
⇒ A + B = 180° - C
Sum of angles of a triangle = 180°
A + B + C = 180°
⇒ A + B = 180° - C
Multiplying both sides by
$\frac{1}{2}$
$=\frac{A+B}{2}=\frac{180^{\circ}-C}{2}$
$=\frac{A+B}{2}=\frac{180^{\circ}}{2}-\frac{C}{2}$
$=\frac{A+B}{2}=90^{\circ}-\frac{C}{2}$ …(3)
$=\frac{A+B}{2}=\frac{180^{\circ}-C}{2}$
$=\frac{A+B}{2}=\frac{180^{\circ}}{2}-\frac{C}{2}$
$=\frac{A+B}{2}=90^{\circ}-\frac{C}{2}$ …(3)
Taking LHS
$\cos \frac{A+B}{2}$
$=\cos \left(90^{\circ}-\frac{\mathrm{C}}{2}\right)$ (from eq (3))
$=\sin \frac{C}{2}$ [∵ cos (90° - θ) = sin θ]
=RHS
Hence Proved
Sol :
sin 3A = cos (A-26°) …(i)
We know that
Sin θ = cos (90° - θ)
So, Eq. (i) become
Cos (90° - 3A) = cos (A -26°)
On Equating both the sides, we get
90° - 3A = A – 26°
⇒ -3A - A = -26° -90°
⇒ -4A = -116°
⇒ A = 29°
Sol :
cos(2 θ +54o)= sin θ …(i)
We know that
Sin θ = cos (90° - θ)
So, Eq. (i) become
cos(2 θ +54o) = cos( 90° - θ)
On Equating both the sides, we get
2θ + 54° = 90° - θ
⇒ 2θ + θ = 90° - 54°
⇒ 3θ = 36°
⇒ θ = 12°
tan 3θ = cot (θ + 18°) …(i)
We know that
tan θ = cot (90° - θ)
So, Eq. (i) become
Cot (90° - 3θ) = cot (θ + 18°)
On Equating both the sides, we get
90° - 3θ = θ + 18°
⇒ -3θ - θ = 18° -90°
⇒ -4θ = -72°
⇒ θ = 18°
sec 5θ = cosec (θ-36°) …(i)
We know that
sec θ = cosec (90° - θ)
So, Eq. (i) become
Cosec (90° - 5θ) = cosec (θ -36°)
On Equating both the sides, we get
90° - 5θ = θ -36°
⇒ -5θ - θ = -36° -90°
⇒ -6θ = -126°
⇒ θ = 21°
Sol :
Taking LHS
sin 70° sec 20°
$\Rightarrow \sin 70^{\circ} \times \frac{1}{\cos 20^{\circ}}$
$\Rightarrow \sin 70^{\circ} \times \frac{1}{\sin \left(90^{\circ}-20^{\circ}\right)}$ [∵ cos θ = sin (90° - θ)]
$\Rightarrow \frac{\sin 70^{\circ}}{\sin 70^{\circ}}$
= 1 = RHS
Hence Proved
Sol :
Taking LHS
Sin(90° - θ) tanθ [∵ cos θ = sin (90° - θ)]
⇒ cos θ tan θ
$\Rightarrow \cos \theta \times \frac{\sin \theta}{\cos \theta}$ [∵ tan θ $=\frac{\sin \theta}{\cos \theta}$]
= sin θ = RHS
Hence Proved
Sol :
Taking LHS
Tan 63° tan 27°
⇒ tan 63° cot (90° - 27°) [∵ tan θ = cot (90° - θ)]
⇒ tan 63° cot 63°
$\cos \frac{A+B}{2}$
$=\cos \left(90^{\circ}-\frac{\mathrm{C}}{2}\right)$ (from eq (3))
$=\sin \frac{C}{2}$ [∵ cos (90° - θ) = sin θ]
=RHS
Hence Proved
Question 12 A
If sin 3A = cos(A – 26o), where 3A is an acute angle, find the value of A.
sin 3A = cos (A-26°) …(i)
We know that
Sin θ = cos (90° - θ)
So, Eq. (i) become
Cos (90° - 3A) = cos (A -26°)
On Equating both the sides, we get
90° - 3A = A – 26°
⇒ -3A - A = -26° -90°
⇒ -4A = -116°
⇒ A = 29°
Question 12 B
Find θ if cos(2 θ +54o)=
sin θ, where (2θ +54o) is an acute
angle.
cos(2 θ +54o)= sin θ …(i)
We know that
Sin θ = cos (90° - θ)
So, Eq. (i) become
cos(2 θ +54o) = cos( 90° - θ)
On Equating both the sides, we get
2θ + 54° = 90° - θ
⇒ 2θ + θ = 90° - 54°
⇒ 3θ = 36°
⇒ θ = 12°
Question 12 C
If tan 3 θ =cot (θ +18o),
where 3 θ and θ +18o are
acute angles, find the value of θ.
Sol :tan 3θ = cot (θ + 18°) …(i)
We know that
tan θ = cot (90° - θ)
So, Eq. (i) become
Cot (90° - 3θ) = cot (θ + 18°)
On Equating both the sides, we get
90° - 3θ = θ + 18°
⇒ -3θ - θ = 18° -90°
⇒ -4θ = -72°
⇒ θ = 18°
Question 12 D
If sec 5 θ =cosec (θ -36o),
where 5 θ is an acute angle, find the value of θ.
Sol :sec 5θ = cosec (θ-36°) …(i)
We know that
sec θ = cosec (90° - θ)
So, Eq. (i) become
Cosec (90° - 5θ) = cosec (θ -36°)
On Equating both the sides, we get
90° - 5θ = θ -36°
⇒ -5θ - θ = -36° -90°
⇒ -6θ = -126°
⇒ θ = 21°
Question 13
Prove that :
sin 70o. sec 20o=1Sol :
Taking LHS
sin 70° sec 20°
$\Rightarrow \sin 70^{\circ} \times \frac{1}{\cos 20^{\circ}}$
$\Rightarrow \sin 70^{\circ} \times \frac{1}{\sin \left(90^{\circ}-20^{\circ}\right)}$ [∵ cos θ = sin (90° - θ)]
$\Rightarrow \frac{\sin 70^{\circ}}{\sin 70^{\circ}}$
= 1 = RHS
Hence Proved
Question 14
Prove that :
sin (90o- θ) tan θ=sin θSol :
Taking LHS
Sin(90° - θ) tanθ [∵ cos θ = sin (90° - θ)]
⇒ cos θ tan θ
$\Rightarrow \cos \theta \times \frac{\sin \theta}{\cos \theta}$ [∵ tan θ $=\frac{\sin \theta}{\cos \theta}$]
= sin θ = RHS
Hence Proved
Question 15
Prove that :
tan 63o. tan 27o=1Sol :
Taking LHS
Tan 63° tan 27°
⇒ tan 63° cot (90° - 27°) [∵ tan θ = cot (90° - θ)]
⇒ tan 63° cot 63°
$\Rightarrow \tan 63^{\circ} \times \frac{1}{\tan
63^{\circ}}$ $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
= 1 =RHS
Hence Proved
Sol :
Taking LHS
$=\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1$
$=\frac{\cos \theta \sin \theta}{\frac{\sin \theta}{\cos \theta}}-1$
$=\frac{\cos \theta \sin \theta \times \cos \theta}{\sin \theta}-1$
= cos2 θ – 1
= - sin2 θ [∵ cos2 θ + sin2 θ = 1]
= RHS
Hence Proved
Sol :
Taking LHS = sin 55 ° cos 48°
We know that
cos θ = sin (90° - θ)
Here, θ = 48°
⇒ sin 55° sin (90° - 48°)
⇒ sin 55° sin 42°
We also know that
Sin θ = cos (90° - θ)
Here, θ = 55°
⇒ cos (90° - 55°) sin 42°
⇒ cos 35° sin 42° = RHS
Hence Proved
Sol :
Taking LHS = sin 25o+sin65o
We know that
Sin θ = cos (90° - θ)
Here, θ = 25°
⇒ cos2 (90° - 25°)+ sin2 65°
⇒ cos2 65° + sin2 65°
= 1 [∵ cos2 θ + sin2 θ = 1]
Now, RHS = cos2 63o+cos2 39o
We know that
cos θ = sin (90° - θ)
Here, θ = 39°
⇒ cos2 63° + sin2 (90° - 39°)
⇒ cos2 63°+ sin2 63°
=1 [∵ cos2 θ + sin2 θ = 1]
LHS = RHS
Hence Proved
Sol :
Taking LHS = sin 54o+cos67o
We know that
cos θ = sin (90° - θ)
Here, θ = 67°
⇒ sin 54°+ sin (90° - 67°)
⇒ sin 54°+ sin 23°
We also know that
Sin θ = cos (90° - θ)
Here, θ = 54°
⇒ cos (90° - 54°)+ sin 23°
⇒ cos 36°+ sin 23° = RHS
Hence Proved
Sol :
Taking LHS = cos 27+ sin51o
We know that
cos θ = sin (90° - θ)
Here, θ = 27°
⇒ sin (90° - 27°)+ sin 51°
⇒ sin 63°+ sin 51°
We also know that
Sin θ = cos (90° - θ)
Here, θ = 51°
⇒ sin 63°+ cos (90° - 51°)
⇒ sin 63°+ cos 39° = RHS
Hence Proved
Sol :
Taking LHS= sin240o+sin250o
⇒ cos2 (90° - 40°) + sin2 50° [∵ Sin θ = cos (90° - θ)]
⇒ cos2 50° + sin2 50°
= 1 =RHS [∵ cos2 θ + sin2 θ = 1]
Hence Proved
Sol :
Taking LHS= sin229o + sin261o
⇒ cos2 (90° - 29°) + sin2 61° [∵ Sin θ = cos (90° - θ)]
⇒ cos2 61° + sin2 61°
= 1 =RHS [∵ cos2 θ + sin2 θ = 1]
Hence Proved
Sol :
Taking LHS = sin θ cos (90° - θ) + cos θ sin (90° - θ)
⇒ sin θ × sin θ + cos θ × cos θ [∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]
⇒ cos2 θ + sin2 θ [∵ cos2 θ + sin2 θ = 1]
= 1 = RHS
Hence Proved
Sol :
Taking LHS = cos θ cos ( 90° - θ) + sin θ sin (90° - θ)
⇒ cos θ × sin θ – sinθ × cos θ [∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]
= 0 = RHS
Hence Proved
Sol :
Taking LHS
= sin 42° cos 48° + cos 42° sin 48°
= cos (90° - 42°) cos 48° + sin (90° - 42°) sin 48°
[∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]
= cos 48° cos 48° + sin 48° sin 48°
= cos2 48° + sin2 48°
= 1 [∵ cos2 θ + sin2 θ = 1]
=LHS=RHS
Hence Proved
Sol :
Taking LHS
$=\frac{\cos 20^{\circ}}{\sin 70^{\circ}}+\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}$
$=\frac{\cos 20^{\circ}}{\cos \left(90^{\circ}-70^{\circ}\right)}+\frac{\cos \theta}{\cos \theta}$ [∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]
$=\frac{\cos 20^{\circ}}{\cos 20^{\circ}}+1$
= 1 + 1
= 2 = RHS
Hence Proved
Sol :
Taking LHS
= tan 27° tan 45° tan 63°
=tan (90° - 27°) tan 45° tan 63° [∵ tan θ = cot (90° - θ)]
=cot 63° tan 45° tan 63° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 63^{\circ}} \times \tan 45^{\circ} \times \tan 63^{\circ}$
= tan 45° [∵ tan 45° =1]
=1 =RHS
Hence Proved
Sol :
Taking LHS
= tan 9° tan 27° tan 45° tan 63° tan 81°
=cot(90° - 9°) tan (90° - 27°) tan 45° tan 63° tan 81° [∵ tan θ = cot (90° - θ)]
=cot 81° cot 63° tan 45° tan 63° tan 81° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 81^{\circ}} \times \frac{1}{\tan 63^{\circ}} \times \tan 45^{\circ} \times \tan 63^{\circ} \times \tan 81^{\circ}$
= tan 45° [∵ tan 45° =1]
=1 =RHS
Hence Proved
= cos9°.cos27°.cos63°.cos81°
Sol :
Taking LHS
= sin 9° sin 27° sin 63° sin 81°
= cos (90° - 9°) cos (90° - 27°) cos (90° – 63°) cos (90°- 81°)
= cos 81° cos 63° cos 27° cos 9°
Or cos 9° cos 27° cos 63° cos 81° = RHS
Hence Proved
Sol :
Taking LHS
= tan 7° tan 23° tan 60° tan 67° tan 83°
=cot(90° - 7°) tan (90° - 23°) tan 60° tan 67° tan 83° [∵ tan θ = cot (90° - θ)]
=cot 83° cot 67° tan 60° tan 67° tan 83° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 83^{\circ}} \times \frac{1}{\tan 67^{\circ}} \times \tan 60^{\circ} \times \tan 67^{\circ} \times \tan 83^{\circ}$
= tan 60° [∵ tan 60° =√3]
=√3 =RHS
Sol :
Taking LHS
= tan 15° tan 25° tan 60° tan 65° tan 75°
=cot(90° - 15°) tan (90° - 25°) tan 60° tan 65° tan 75° [∵ tan θ = cot (90° - θ)]
=cot 75° cot 65° tan 60° tan 65° tan 75° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 75^{\circ}} \times \frac{1}{\tan 65^{\circ}} \times \tan 60^{\circ} \times \tan 65^{\circ} \times \tan 75^{\circ}$
= tan 60° [∵ tan 60° =√3]
=√3 =RHS
Sol :
$=\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}$
= 1 =RHS
Hence Proved
Question 16
Prove that :
$\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1=-\sin ^{2} \theta$Sol :
Taking LHS
$=\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1$
$=\frac{\cos \theta \sin \theta}{\frac{\sin \theta}{\cos \theta}}-1$
$=\frac{\cos \theta \sin \theta \times \cos \theta}{\sin \theta}-1$
= cos2 θ – 1
= - sin2 θ [∵ cos2 θ + sin2 θ = 1]
= RHS
Hence Proved
Question 17
Prove that :
sin 55o. cos 48o=cos35o. sin 42oTaking LHS = sin 55 ° cos 48°
We know that
cos θ = sin (90° - θ)
Here, θ = 48°
⇒ sin 55° sin (90° - 48°)
⇒ sin 55° sin 42°
We also know that
Sin θ = cos (90° - θ)
Here, θ = 55°
⇒ cos (90° - 55°) sin 42°
⇒ cos 35° sin 42° = RHS
Hence Proved
Question 18
Prove that :
sin2 25o+sin2 65° =
cos2 63°+cos2 39oSol :
Taking LHS = sin 25o+sin65o
We know that
Sin θ = cos (90° - θ)
Here, θ = 25°
⇒ cos2 (90° - 25°)+ sin2 65°
⇒ cos2 65° + sin2 65°
= 1 [∵ cos2 θ + sin2 θ = 1]
Now, RHS = cos2 63o+cos2 39o
We know that
cos θ = sin (90° - θ)
Here, θ = 39°
⇒ cos2 63° + sin2 (90° - 39°)
⇒ cos2 63°+ sin2 63°
=1 [∵ cos2 θ + sin2 θ = 1]
LHS = RHS
Hence Proved
Question 19
Prove that :
sin 54o+cos67o= sin23o+cos36oSol :
Taking LHS = sin 54o+cos67o
We know that
cos θ = sin (90° - θ)
Here, θ = 67°
⇒ sin 54°+ sin (90° - 67°)
⇒ sin 54°+ sin 23°
We also know that
Sin θ = cos (90° - θ)
Here, θ = 54°
⇒ cos (90° - 54°)+ sin 23°
⇒ cos 36°+ sin 23° = RHS
Hence Proved
Question 20
Prove that :
cos 27+ sin51o = sin63o+cos 39oSol :
Taking LHS = cos 27+ sin51o
We know that
cos θ = sin (90° - θ)
Here, θ = 27°
⇒ sin (90° - 27°)+ sin 51°
⇒ sin 63°+ sin 51°
We also know that
Sin θ = cos (90° - θ)
Here, θ = 51°
⇒ sin 63°+ cos (90° - 51°)
⇒ sin 63°+ cos 39° = RHS
Hence Proved
Question 21
Prove that :
sin240o+sin250o=1Sol :
Taking LHS= sin240o+sin250o
⇒ cos2 (90° - 40°) + sin2 50° [∵ Sin θ = cos (90° - θ)]
⇒ cos2 50° + sin2 50°
= 1 =RHS [∵ cos2 θ + sin2 θ = 1]
Hence Proved
Question 22
Prove that :
sin229o + sin261o=1Sol :
Taking LHS= sin229o + sin261o
⇒ cos2 (90° - 29°) + sin2 61° [∵ Sin θ = cos (90° - θ)]
⇒ cos2 61° + sin2 61°
= 1 =RHS [∵ cos2 θ + sin2 θ = 1]
Hence Proved
Question 23
Prove that :
sin θ .cos (90° - θ) + cos θ sin (90° - θ).Sol :
Taking LHS = sin θ cos (90° - θ) + cos θ sin (90° - θ)
⇒ sin θ × sin θ + cos θ × cos θ [∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]
⇒ cos2 θ + sin2 θ [∵ cos2 θ + sin2 θ = 1]
= 1 = RHS
Hence Proved
Question 24
Prove that :
cos θ . cos(90° – θ) + sin θ sin (90° – θ) = 0Sol :
Taking LHS = cos θ cos ( 90° - θ) + sin θ sin (90° - θ)
⇒ cos θ × sin θ – sinθ × cos θ [∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]
= 0 = RHS
Hence Proved
Question 25
Prove that :
sin 42°. cos 48° + cos 42° . sin 48° = 1Sol :
Taking LHS
= sin 42° cos 48° + cos 42° sin 48°
= cos (90° - 42°) cos 48° + sin (90° - 42°) sin 48°
[∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]
= cos 48° cos 48° + sin 48° sin 48°
= cos2 48° + sin2 48°
= 1 [∵ cos2 θ + sin2 θ = 1]
=LHS=RHS
Hence Proved
Question 26
Prove that :
$\frac{\cos 20^{\circ}}{\sin 70^{\circ}}+\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}=2$Sol :
Taking LHS
$=\frac{\cos 20^{\circ}}{\sin 70^{\circ}}+\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}$
$=\frac{\cos 20^{\circ}}{\cos \left(90^{\circ}-70^{\circ}\right)}+\frac{\cos \theta}{\cos \theta}$ [∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]
$=\frac{\cos 20^{\circ}}{\cos 20^{\circ}}+1$
= 1 + 1
= 2 = RHS
Hence Proved
Question 27
Prove that :
tan 27° tan 45° tan 63°Sol :
Taking LHS
= tan 27° tan 45° tan 63°
=tan (90° - 27°) tan 45° tan 63° [∵ tan θ = cot (90° - θ)]
=cot 63° tan 45° tan 63° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 63^{\circ}} \times \tan 45^{\circ} \times \tan 63^{\circ}$
= tan 45° [∵ tan 45° =1]
=1 =RHS
Hence Proved
Question 28
Prove that :
tan 9°. tan 27°. tan 45°. tan 63°. tan 81° = 1Sol :
Taking LHS
= tan 9° tan 27° tan 45° tan 63° tan 81°
=cot(90° - 9°) tan (90° - 27°) tan 45° tan 63° tan 81° [∵ tan θ = cot (90° - θ)]
=cot 81° cot 63° tan 45° tan 63° tan 81° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 81^{\circ}} \times \frac{1}{\tan 63^{\circ}} \times \tan 45^{\circ} \times \tan 63^{\circ} \times \tan 81^{\circ}$
= tan 45° [∵ tan 45° =1]
=1 =RHS
Hence Proved
Question 29
Prove that :
sin 9°. sin 27°. sin 63°. sin 81°= cos9°.cos27°.cos63°.cos81°
Sol :
Taking LHS
= sin 9° sin 27° sin 63° sin 81°
= cos (90° - 9°) cos (90° - 27°) cos (90° – 63°) cos (90°- 81°)
= cos 81° cos 63° cos 27° cos 9°
Or cos 9° cos 27° cos 63° cos 81° = RHS
Hence Proved
Question 30 A
Prove that :
$\tan 7^{\circ} \cdot \tan 23^{\circ} \cdot \tan 60^{\circ} \cdot \tan 67^{\circ} \cdot \tan
83^{\circ}=\sqrt{3}$Sol :
Taking LHS
= tan 7° tan 23° tan 60° tan 67° tan 83°
=cot(90° - 7°) tan (90° - 23°) tan 60° tan 67° tan 83° [∵ tan θ = cot (90° - θ)]
=cot 83° cot 67° tan 60° tan 67° tan 83° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 83^{\circ}} \times \frac{1}{\tan 67^{\circ}} \times \tan 60^{\circ} \times \tan 67^{\circ} \times \tan 83^{\circ}$
= tan 60° [∵ tan 60° =√3]
=√3 =RHS
Question 30 B
Prove that :
$\tan 15^{\circ} \tan 25^{\circ} \tan 60^{\circ} \tan 65^{\circ} \tan 75^{\circ}=\sqrt{3}$Sol :
Taking LHS
= tan 15° tan 25° tan 60° tan 65° tan 75°
=cot(90° - 15°) tan (90° - 25°) tan 60° tan 65° tan 75° [∵ tan θ = cot (90° - θ)]
=cot 75° cot 65° tan 60° tan 65° tan 75° $\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$
$=\frac{1}{\tan 75^{\circ}} \times \frac{1}{\tan 65^{\circ}} \times \tan 60^{\circ} \times \tan 65^{\circ} \times \tan 75^{\circ}$
= tan 60° [∵ tan 60° =√3]
=√3 =RHS
Question 31
Find the value off the following:
$\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\cos \mathrm{ec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ}
\cdot \cos \mathrm{ec} 40^{\circ}$Sol :
$=\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}$
$=\frac{\sin 50^{\circ}}{\sin \left(90^{\circ}-40^{\circ}\right)}+\frac{\operatorname{cosec} 40^{\circ}}{\operatorname{cosec}\left(90^{\circ}-50^{\circ}\right)}-4 \sin \left(90^{\circ}-50^{\circ}\right) \operatorname{cosec} 40^{\circ}$
[∵ cos θ = sin (90° - θ) and sec θ = cosec (90° - θ)]
$=\frac{\sin 50^{\circ}}{\sin
50^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\operatorname{cosec} 40^{\circ}}-4 \sin 40^{\circ} \times
\frac{1}{\sin 40^{\circ}}$ $\left[\because \sin \theta=\frac{1}{\operatorname{cosec} \theta}\right]$
= 1 + 1 – 4
= -2
Sol :
$=\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2} 31^{\circ}}+\cos \left(90^{\circ}-35^{\circ}\right) \sec 55^{\circ}$
= 1 + 1 – 4
= -2
Question 32
Find the value off the following:
$\frac{\cos ^{2} 20^{0}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2} 31^{\circ}}+\sin 35^{\circ} \cdot \sec
55^{\circ}$Sol :
$=\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2} 31^{\circ}}+\cos \left(90^{\circ}-35^{\circ}\right) \sec 55^{\circ}$
$=\frac{\cos ^{2} 20^{\circ}+\sin
^{2}\left(90^{\circ}-70^{\circ}\right)}{\sin ^{2}\left(59^{\circ}\right)+\cos
^{2}\left(90^{\circ}-31^{\circ}\right)}+\cos 55^{\circ} \sec 55^{\circ}$
[∵ cos θ = sin (90° - θ) and sec θ = cosec (90° - θ)]
[∵ cos θ = sin (90° - θ) and sec θ = cosec (90° - θ)]
$=\frac{\cos ^{2} 20^{\circ}+\sin ^{2} 20^{\circ}}{\sin
^{2}\left(59^{\circ}\right)+\cos ^{2} 59^{\circ}}+\cos 55^{\circ} \times \frac{1}{\cos 55^{\circ}}$
[∵ cos2 θ + sin2 θ =
1]
= 1 + 1
=2
Sol :
$=\frac{\tan 50^{\circ}+\sec 50^{\circ}}{\cot 40^{\circ}+\operatorname{cosec} 40^{\circ}}+\cos 40^{\circ} \operatorname{cosec} 50^{\circ}$
=2
Question 33
Find the value off the following:
$\frac{\tan 50^{\circ}+\sec 50^{\circ}}{\cot 40^{\circ}+\cos \mathrm{ec} 40^{\circ}}+\cos 40^{\circ} \cdot
\operatorname{cosec} 50^{\circ}$Sol :
$=\frac{\tan 50^{\circ}+\sec 50^{\circ}}{\cot 40^{\circ}+\operatorname{cosec} 40^{\circ}}+\cos 40^{\circ} \operatorname{cosec} 50^{\circ}$
$=\frac{\cot
\left(90^{\circ}-50^{\circ}\right)+\operatorname{cosec}\left(90^{\circ}-50^{\circ}\right)}{\cot
40^{\circ}+\operatorname{cosec} 40^{\circ}}+\sin \left(90^{\circ}-40^{\circ}\right) \operatorname{cosec}
50^{\circ}$
[∵ tan θ = cot (90° - θ) , sec θ = cosec (90° - θ) and cos θ = sin (90° - θ)]
[∵ tan θ = cot (90° - θ) , sec θ = cosec (90° - θ) and cos θ = sin (90° - θ)]
$=\frac{\cot 40^{\circ}+\operatorname{cosec} 40^{\circ}}{\cot
40^{\circ}+\operatorname{cosec} 40^{\circ}}+\sin 50^{\circ} \operatorname{cosec} 50^{\circ}$
$=1+\sin 50^{\circ} \times \frac{1}{\sin 50^{\circ}}$ $\because \sin \theta=\frac{1}{\operatorname{cosec}
\theta}$
= 1 + 1
= 2
Sol :
cosec (65° + θ) – sec (25° - θ) – tan(55° - θ) + cot(35° +θ)
= sec {90°-(65°+θ)} – sec (25° - θ) – tan(55° - θ) + tan {90°-(35° +θ)}
[∵ cosec θ = sec (90° - θ) and cot θ = tan (90° - θ)]
= sec ( 90° - 65°-θ) – sec (25° - θ) – tan(55° - θ) + tan (90°- 35° - θ)
= sec (25° - θ) - sec (25° - θ) – tan(55° - θ) + tan (55° - θ)
= 0
Sol :
$=\frac{\cos 35^{\circ}}{\sin 55^{\circ}}+\frac{\sin 11^{\circ}}{\cos 79^{\circ}}-\cos 28^{\circ} \operatorname{cosec} 62^{\circ}$
= 2
Question 34
Find the value off the following:
cosec (65° + θ) – sec (25° - θ) – tan(55° - θ) + cot(35° +θ)Sol :
cosec (65° + θ) – sec (25° - θ) – tan(55° - θ) + cot(35° +θ)
= sec {90°-(65°+θ)} – sec (25° - θ) – tan(55° - θ) + tan {90°-(35° +θ)}
[∵ cosec θ = sec (90° - θ) and cot θ = tan (90° - θ)]
= sec ( 90° - 65°-θ) – sec (25° - θ) – tan(55° - θ) + tan (90°- 35° - θ)
= sec (25° - θ) - sec (25° - θ) – tan(55° - θ) + tan (55° - θ)
= 0
Question 35
Find the value off the following:
$\frac{\cos 35^{\circ}}{\sin 55^{\circ}}+\frac{\sin 11^{\circ}}{\cos 79^{\circ}}-\cos 28^{\circ} \cdot
\operatorname{cosec} 62^{\circ}$Sol :
$=\frac{\cos 35^{\circ}}{\sin 55^{\circ}}+\frac{\sin 11^{\circ}}{\cos 79^{\circ}}-\cos 28^{\circ} \operatorname{cosec} 62^{\circ}$
$=\frac{\sin \left(90^{\circ}-35^{\circ}\right)}{\sin
55^{\circ}}+\frac{\sin 11^{\circ}}{\sin \left(90^{\circ}-79^{\circ}\right)}-\sin \left(90^{\circ}-28^{\circ}\right)
\operatorname{cosec} 62^{\circ}$
$=\frac{\sin 55^{\circ}}{\sin 55^{\circ}}+\frac{\sin
11^{\circ}}{\sin 11^{\circ}}-\sin 62^{\circ} \operatorname{cosec} 62^{\circ}$
$=1+1-\sin 62^{\circ} \times \frac{1}{\sin 62^{\circ}}$
= 1 + 1 – 1
= 1
Sol :
$=\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2} 31^{\circ}}$
= 1 + 1 – 1
= 1
Question 36
Find the value off the following:
$\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2} 31^{\circ}}$Sol :
$=\frac{\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}}{\sin ^{2} 59^{\circ}+\sin ^{2} 31^{\circ}}$
$=\frac{\cos ^{2} 20^{\circ}+\sin
^{2}\left(90^{\circ}-70^{\circ}\right)}{\sin ^{2}\left(59^{\circ}\right)+\cos
^{2}\left(90^{\circ}-31^{\circ}\right)}$
[∵ cos θ = sin (90° - θ) and sec θ = cosec (90° - θ)]
$=\frac{\cos ^{2} 20^{\circ}+\sin ^{2} 20^{\circ}}{\sin ^{2}\left(59^{\circ}\right)+\cos ^{2} 59^{\circ}}$
= 1
Sol :
cosec (65° + θ) – sec (25° - θ)
= sec {90°-(65°+θ)} – sec (25° - θ)
[∵ cosec θ = sec (90° - θ)]
= sec ( 90° - 65°-θ) – sec (25° - θ)
= sec (25° - θ) - sec (25° - θ)
= 0
Sol :
cos (60° + θ) – sin (30° - θ)
= sin {90°-(60°+θ)} – sin (30° - θ) [∵ cos θ = sin (90° - θ)]
= sin ( 90° - 60°-θ) – sin (30° - θ)
= sin (30° - θ) - sin (30° - θ)
= 0
Sol :
sec 70° sin 20° - cos 20° cosec 70°
= cosec (90°-70°) cos (90° - 20°)- cos 20° cosec 70°
[∵ sec θ = cosec (90° - θ) and Sin θ = cos (90° - θ)]
= cosec 70° cos 20° - cos 20° cosec 70°
=0
Sol :
(sin 72° + cos 18°)( sin 72° - cos 18°)
Using the identity , (a-b)(a+b) = a2 – b2
= (sin 72°)2 - (cos 18°)2
= {cos(90° - 72°)}2 - (cos 18°)2 [∵ Sin θ = cos (90° - θ)]
=(cos 18°)2 - (cos 18°)2
= 0
Sol :
$=\left(\frac{\sin 35^{\circ}}{\cos 55^{\circ}}\right)^{2}+\left(\frac{\cos 55^{\circ}}{\sin 35^{\circ}}\right)^{2}-2 \cos 60^{\circ}$
$=\left(\frac{\sin 35^{\circ}}{\sin \left(90^{\circ}-55^{\circ}\right)}\right)^{2}+\left(\frac{\sin \left(90^{\circ}-55^{\circ}\right)}{\sin 35^{\circ}}\right)^{2}-2 \cos 60^{\circ}$ [∵ cos θ = sin (90° - θ)]
[∵ cos θ = sin (90° - θ) and sec θ = cosec (90° - θ)]
$=\frac{\cos ^{2} 20^{\circ}+\sin ^{2} 20^{\circ}}{\sin ^{2}\left(59^{\circ}\right)+\cos ^{2} 59^{\circ}}$
= 1
Question 37
Find the value off the following:
cosec (65° + θ) – sec (25° - θ)Sol :
cosec (65° + θ) – sec (25° - θ)
= sec {90°-(65°+θ)} – sec (25° - θ)
[∵ cosec θ = sec (90° - θ)]
= sec ( 90° - 65°-θ) – sec (25° - θ)
= sec (25° - θ) - sec (25° - θ)
= 0
Question 38
Find the value off the following:
cos (60° + θ) – sin (30° - θ)Sol :
cos (60° + θ) – sin (30° - θ)
= sin {90°-(60°+θ)} – sin (30° - θ) [∵ cos θ = sin (90° - θ)]
= sin ( 90° - 60°-θ) – sin (30° - θ)
= sin (30° - θ) - sin (30° - θ)
= 0
Question 39
Find the value off the following:
sec 70°. sin 20° - cos 20°. cosec 70°Sol :
sec 70° sin 20° - cos 20° cosec 70°
= cosec (90°-70°) cos (90° - 20°)- cos 20° cosec 70°
[∵ sec θ = cosec (90° - θ) and Sin θ = cos (90° - θ)]
= cosec 70° cos 20° - cos 20° cosec 70°
=0
Question 40
Find the value off the following:
(sin 72° + cos 18°)( sin 72° - cos 18°)Sol :
(sin 72° + cos 18°)( sin 72° - cos 18°)
Using the identity , (a-b)(a+b) = a2 – b2
= (sin 72°)2 - (cos 18°)2
= {cos(90° - 72°)}2 - (cos 18°)2 [∵ Sin θ = cos (90° - θ)]
=(cos 18°)2 - (cos 18°)2
= 0
Question 41
Find the value off the following:
$\left(\frac{\sin 35^{\circ}}{\cos 55^{\circ}}\right)^{2}+\left(\frac{\cos 55^{\circ}}{\sin 35^{\circ}}\right)-2
\cos 60^{\circ}$Sol :
$=\left(\frac{\sin 35^{\circ}}{\cos 55^{\circ}}\right)^{2}+\left(\frac{\cos 55^{\circ}}{\sin 35^{\circ}}\right)^{2}-2 \cos 60^{\circ}$
$=\left(\frac{\sin 35^{\circ}}{\sin \left(90^{\circ}-55^{\circ}\right)}\right)^{2}+\left(\frac{\sin \left(90^{\circ}-55^{\circ}\right)}{\sin 35^{\circ}}\right)^{2}-2 \cos 60^{\circ}$ [∵ cos θ = sin (90° - θ)]
$=\left(\frac{\sin 35^{\circ}}{\sin
35^{\circ}}\right)^{2}+\left(\frac{\sin 35^{\circ}}{\sin 35^{\circ}}\right)^{2}-2\left(\frac{1}{2}\right)$
= 1 + 1 - 1
= 1
Sol :
$=\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}$
= 1 + 1 - 1
= 1
Question 42
Find the value off the following:
$\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \cdot \operatorname{cosec} 31^{\circ}$Sol :
$=\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}$
$=\frac{\sin \left(90^{\circ}-80^{\circ}\right)}{\sin
10^{\circ}}+\sin \left(90^{\circ}-59^{\circ}\right) \operatorname{cosec} 31^{\circ}$
$=\frac{\sin 10^{\circ}}{\sin 10^{\circ}}+\sin 31^{\circ}
\operatorname{cosec} 31^{\circ}$
$=1+\sin 31^{\circ} \times \frac{1}{\sin 31^{\circ}}$
= 1 + 1
= 2
Sol :
(sin 50° + θ)- cos (40° -θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°
= 1 + 1
= 2
Question 43
Find the value off the following:
(sin 50° + θ)- cos (40° - θ) + tan 1°. tan 10° tan 20°. tan 70°. tan 80°. tan 89°Sol :
(sin 50° + θ)- cos (40° -θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°
= cos {90° -(50° + θ)} - cos (40° -θ)+ cot(90° - 1°) tan (90° - 10°) cot(90° - 20°) tan 70° tan 80° tan 89°
[∵ Sin θ = cos (90° - θ) & tan θ = cot (90° - θ)]
= cos (40° -θ) - cos (40° -θ) + cot89° cot80° cot70° tan 70° tan 80° tan 89°
$=\frac{1}{\tan 89^{\circ}} \times \frac{1}{\tan 80^{\circ}} \times
\frac{1}{\tan 70^{\circ}} \times \tan 70^{\circ} \tan 80^{\circ} \tan 89^{\circ}$ $\left[\because \cot
\theta=\frac{1}{\tan \theta}\right]$
= 1
Sol :
$=\sec ^{2} 10^{\circ}-\cot ^{2} 80^{\circ}+\frac{\sin 15^{\circ} \cos 75^{\circ}+\cos 15^{\circ} \sin 75^{\circ}}{\cos \theta \sin \left(90^{\circ}-\theta\right)+\sin \theta \cos \left(90^{\circ}-\theta\right)}$
= 1
Question 44
Find the value off the following:
$\sec ^{2} 10^{\circ}-\cot ^{2} 80^{\circ}+\frac{\sin 15^{\circ} \cos 75^{\circ}+\cos 15^{\circ} \cdot \sin
75^{\circ}}{\cos \theta \sin (90-\theta)+\sin \theta \cos \left(90^{\circ}-\theta\right)}$Sol :
$=\sec ^{2} 10^{\circ}-\cot ^{2} 80^{\circ}+\frac{\sin 15^{\circ} \cos 75^{\circ}+\cos 15^{\circ} \sin 75^{\circ}}{\cos \theta \sin \left(90^{\circ}-\theta\right)+\sin \theta \cos \left(90^{\circ}-\theta\right)}$
$=\sec ^{2} 10^{\circ}-\tan
^{2}\left(90^{\circ}-80^{\circ}\right)+\frac{\cos \left(90^{\circ}-15^{\circ}\right) \cos 75^{\circ}+\sin
\left(90^{\circ}-15^{\circ}\right) \sin 75^{\circ}}{\cos \theta \sin \left(90^{\circ}-\theta\right)+\sin \theta \cos
\left(90^{\circ}-\theta\right)}$
[∵ cot θ = tan (90° - θ), cos θ = sin (90° - θ) and Sin θ = cos (90° - θ)]
[∵ cot θ = tan (90° - θ), cos θ = sin (90° - θ) and Sin θ = cos (90° - θ)]
$=\sec ^{2} 10^{\circ}-\tan ^{2} 10^{\circ}+\frac{\cos 75^{\circ}
\cos 75^{\circ}+\sin 75^{\circ} \sin 75^{\circ}}{\cos \theta \cos (\theta)+\sin \theta \sin \theta}$
[∵ 1+tan2 θ =sec2 θ and cos2 θ +sin2θ = 1]
[∵ 1+tan2 θ =sec2 θ and cos2 θ +sin2θ = 1]
$=1+\frac{\cos ^{2} 75^{\circ}+\sin ^{2} 75^{\circ}}{\cos ^{2}
\theta+\sin ^{2} \theta}$
= 1 + 1
= 2
Sol :
$\cos \left(40^{\circ}+\theta\right)-\sin \left(50^{\circ}-\theta\right)+\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}$
= 1 + 1
= 2
Question 45
Find the value off the following:
$\cos \left(40^{\circ}+\theta\right)-\sin \left(50^{\circ}-\theta\right)+\frac{\cos ^{2} 40^{\circ}+\cos ^{2}
50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}$Sol :
$\cos \left(40^{\circ}+\theta\right)-\sin \left(50^{\circ}-\theta\right)+\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}$
$=\sin
\left\{90^{\circ}-\left(40^{\circ}+\theta\right)\right\}-\sin \left(50^{\circ}-\theta\right)+\frac{\cos ^{2}
40^{\circ}+\sin ^{2}\left(90^{\circ}-50^{\circ}\right)}{\sin ^{2} 40^{\circ}+\cos
^{2}\left(90^{\circ}-50^{\circ}\right)}$
[∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ)]
$=\sin \left(50^{\circ}-\theta\right)-\sin
\left(50^{\circ}-\theta\right)+\frac{\cos ^{2} 40^{\circ}+\sin ^{2}\left(40^{\circ}\right)}{\sin ^{2} 40^{\circ}+\cos
^{2}\left(40^{\circ}\right)}$
= 0 + 1
= 1
Sol :
$=\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 55^{\circ} \operatorname{cosec} 35^{\circ}}{\tan 5^{\circ} \tan 25^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}}$
= 0 + 1
= 1
Question 46
Find the value off the following:
$=\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 55^{\circ}, \cos \mathrm{ec} 35^{\circ}}{\tan 5^{\circ} \cdot
\tan 25^{\circ} \cdot \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}}$Sol :
$=\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 55^{\circ} \operatorname{cosec} 35^{\circ}}{\tan 5^{\circ} \tan 25^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}}$
$=\left(\frac{\sin \left(90^{\circ}-70^{\circ}\right)}{\sin
20^{\circ}}\right)+\frac{\sin \left(90^{\circ}-55^{\circ}\right) \operatorname{cosec} 35^{\circ}}{\cot
\left(90^{\circ}-5^{\circ}\right) \cot \left(90^{\circ}-25^{\circ}\right) \tan 45^{\circ} \tan 65^{\circ} \tan
85^{\circ}}$
[∵ cos θ = sin (90° - θ) and tan θ = cot (90° - θ)]
[∵ cos θ = sin (90° - θ) and tan θ = cot (90° - θ)]
$=\frac{\sin 20^{\circ}}{\sin 20^{\circ}}+\frac{\sin 35^{\circ}
\operatorname{cosec} 35^{\circ}}{\cot 85^{\circ} \cot 65^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan
85^{\circ}}$
$=1+\frac{\sin 35^{\circ} \times \frac{1}{\sin
35^{\circ}}}{\frac{1}{\tan 85^{\circ}} \times \frac{1}{\tan 65^{\circ}} \times \tan 45^{\circ} \times \tan 65^{\circ}
\times \tan 85^{\circ}}$
= 1 + 1 [∵ tan 45° = 1]
= 2
Sol :
$=\left(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\right)^{2}+\left(\frac{\cos 63^{\circ}}{\sin 27^{\circ}}\right)^{2}$
$=\left(\frac{\sin 27^{\circ}}{\sin \left(90^{\circ}-63^{\circ}\right)}\right)^{2}+\left(\frac{\sin \left(90^{\circ}-63^{\circ}\right)}{\sin 27^{\circ}}\right)^{2}$ [∵ cos θ = sin (90° - θ)]
$=\left(\frac{\sin 27^{\circ}}{\sin 27^{\circ}}\right)^{2}+\left(\frac{\sin 27^{\circ}}{\sin 27^{\circ}}\right)^{2}$
= 1 + 1
= 2
Sol :
$=\frac{3 \sin 5^{\circ}}{\cos 85^{\circ}}+\frac{2 \cos 33^{\circ}}{\sin 57^{\circ}}$
$=\frac{3 \sin 5^{\circ}}{\sin \left(90^{\circ}-85^{\circ}\right)}+\frac{2 \sin \left(90^{\circ}-33^{\circ}\right)}{\sin 57^{\circ}}$ [∵ cos θ = sin (90° - θ)]
$=\frac{3 \sin 5^{\circ}}{\sin 5^{\circ}}+\frac{2 \sin 57^{\circ}}{\sin 57^{\circ}}$
= 3 + 2
= 5
Sol :
$\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2$
$=\frac{\cot 54^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}+\frac{\cot \left(90^{\circ}-20^{\circ}\right)}{\cot 70^{\circ}}-2$ [∵ tan θ = cot (90° - θ)]
$=\frac{\cot 54^{\circ}}{\cot 54^{\circ}}+\frac{\cot 70^{\circ}}{\cot 70^{\circ}}-2$
= 1 +1 – 2
= 0
Sol :
$\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}$
= 1 + 1 [∵ tan 45° = 1]
= 2
Question 47
Find the value off the following:
$\left(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\right)^{2}+\left(\frac{\cos 63^{\circ}}{\sin
27^{\circ}}\right)^{2}$Sol :
$=\left(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\right)^{2}+\left(\frac{\cos 63^{\circ}}{\sin 27^{\circ}}\right)^{2}$
$=\left(\frac{\sin 27^{\circ}}{\sin \left(90^{\circ}-63^{\circ}\right)}\right)^{2}+\left(\frac{\sin \left(90^{\circ}-63^{\circ}\right)}{\sin 27^{\circ}}\right)^{2}$ [∵ cos θ = sin (90° - θ)]
$=\left(\frac{\sin 27^{\circ}}{\sin 27^{\circ}}\right)^{2}+\left(\frac{\sin 27^{\circ}}{\sin 27^{\circ}}\right)^{2}$
= 1 + 1
= 2
Question 48 A
Evaluate the following
$\frac{3 \sin 5^{\circ}}{\cos 85^{\circ}}+\frac{2 \cos 33^{\circ}}{\sin 57^{\circ}}$Sol :
$=\frac{3 \sin 5^{\circ}}{\cos 85^{\circ}}+\frac{2 \cos 33^{\circ}}{\sin 57^{\circ}}$
$=\frac{3 \sin 5^{\circ}}{\sin \left(90^{\circ}-85^{\circ}\right)}+\frac{2 \sin \left(90^{\circ}-33^{\circ}\right)}{\sin 57^{\circ}}$ [∵ cos θ = sin (90° - θ)]
$=\frac{3 \sin 5^{\circ}}{\sin 5^{\circ}}+\frac{2 \sin 57^{\circ}}{\sin 57^{\circ}}$
= 3 + 2
= 5
Question 48 B
Evaluate the following
$\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2$Sol :
$\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2$
$=\frac{\cot 54^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}+\frac{\cot \left(90^{\circ}-20^{\circ}\right)}{\cot 70^{\circ}}-2$ [∵ tan θ = cot (90° - θ)]
$=\frac{\cot 54^{\circ}}{\cot 54^{\circ}}+\frac{\cot 70^{\circ}}{\cot 70^{\circ}}-2$
= 1 +1 – 2
= 0
Question 48 C
Evaluate the following
$\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}$Sol :
$\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}$
$=\frac{\cos 80^{\circ}}{\cos \left(90^{\circ}-10^{\circ}\right)}+\sin \left(90^{\circ}-59^{\circ}\right) \operatorname{cosec} 31^{\circ}$
[∵ cos θ = sin (90° - θ) and sec θ = cosec (90° - θ)]
$=\frac{\cos 80^{\circ}}{\cos 80^{\circ}}+\sin 31^{\circ}
\operatorname{cosec} 31^{\circ}$
$=1+\sin 31^{\circ} \times \frac{1}{\sin 31^{\circ}}$ $\because \sin \theta=\frac{1}{\operatorname{cosec} \theta}$
$=1+\sin 31^{\circ} \times \frac{1}{\sin 31^{\circ}}$ $\because \sin \theta=\frac{1}{\operatorname{cosec} \theta}$
= 1 + 1
= 2
Sol :
We know that
cos θ = sin (90° - θ)
=sin (90° - 38°) sin (90° -52°) – sin 38° sin 52°
= sin 52° sin 38°– sin 38° sin 52°
=0
Sol :
We know that
sec θ = cosec (90° - θ) and cos θ = sin (90° - θ)
Cosec (90° – 41°) sin 49° + sin (90° – 49°) cosec 41°
= cosec 49° sin 49° + sin 41° cosec 41°
= 2
Question 48 D
Evaluate the following
cos38° cos52° – sin38° sin 52°Sol :
We know that
cos θ = sin (90° - θ)
=sin (90° - 38°) sin (90° -52°) – sin 38° sin 52°
= sin 52° sin 38°– sin 38° sin 52°
=0
Question 48 E
Evaluate the following
sec41° sin49° + cos49° cosec 41°Sol :
We know that
sec θ = cosec (90° - θ) and cos θ = sin (90° - θ)
Cosec (90° – 41°) sin 49° + sin (90° – 49°) cosec 41°
= cosec 49° sin 49° + sin 41° cosec 41°
$=\frac{1}{\sin 49^{\circ}} \times \sin 49^{\circ}+\sin 41^{\circ}
\times \frac{1}{\sin 41^{\circ}}$ $\left[\because \sin \theta=\frac{1}{\operatorname{cosec} \theta}\right]$
= 1+ 1
= 2
= 1+ 1
= 2
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