OP Malhotra Class 11 Maths Solutions Chapter 5 Compound and Multiple Angles Chapter Test

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S Chand Class 11 ICSE Maths Solutions Chapter 5 Compound and Multiple Angles Chapter Test

Question 1.
Show that tan 75° = \(\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}\). Hence deduce that tan 75° – cot 75° = 4 sin 60°.
Solution:

Question 2.
Prove that sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2)x = cos x.
Solution:
L.H.S. = sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x
= cos [(n + 2)x – (n + 1) x] = cos (2x – x) = cos x = R.H.S (∵ sin A sin B + cos A cos B = cos (A – B)]

Question 3.
If A + B + C = π, and cos A = cos B cos C, show that 2 cot B cot C = 1.
Solution:
Given A + B + C = π
cos A = cos B cos C ⇒ cos [π – \(\overline{\mathrm{B}+\mathrm{C}}\) ] = cos B cos C
⇒ – cos (B + C) = cos B cos C ⇒ – cos B cos C + sin B sin C = cos B cos C
⇒ 2 cos B cos C = sin B sin C ⇒ \(\frac{2 \cos B \cos C}{\sin B \sin C}\) = 1 ⇒ cot B cot C = 1

Question 4.
Show that \(\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}\) = 3, given that tan α = 2 tan ß.
Solution:

Question 5.
Show that \(\frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}}\) = tan 55°.
Solution:

Question 6.
If sin 2A = \(\frac { 4 }{ 5 }\), find the value of tan A, (0° ≤ A ≤ \(\frac { π }{ 4 }\))
Solution:

Question 7.
Express (i) cot A in terms of cos 2 A, (ii) cos 4θ in terms of cos θ.
Solution:
(i) cos² A = \(\frac{1+\cos 2 \mathrm{~A}}{2}\) and sin² A = \(\frac{1-\cos 2 \mathrm{~A}}{2}\)
∴ cot² A = \(\frac{1+\cos 2 A}{1-\cos 2 A}\) cot A = ± \(\sqrt{\frac{1+\cos 2 A}{1-\cos 2 A}}\)

(ii) cos 4 θ = cos (2 x 2θ) = 2 cos² 2θ – 1 = 2 [2 cos²θ – 1]² – 1
= 2 [4 cos⁴ θ – 4 cos² θ + 1] – 1 = 8 cos⁴ θ – 8 cos² θ + 1

Question 8.
A positive acute angle is divided into two parts whose tangents are \(\frac { 1 }{ 2 }\) and \(\frac { 1 }{ 3 }\). Show that the angle is \(\frac { π }{ 4 }\).
Solution:
Let θ be the positive acute angle s.t θ = α + ß
Now tan θ = tan (α + ß) = \(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}=\frac{\frac{5}{6}}{1-\frac{1}{6}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1 \Rightarrow \theta=\frac{\pi}{4}\)

Question 9.
Show that cos 10° + cos 110° + cos 130° = 0.
Solution:

Question 10.
Show that \(\frac{\sin 5 A+2 \sin 8 A+\sin 11 A}{\sin 8 A+2 \sin 11 A+\sin 14 A}=\frac{\sin 8 A}{\sin 11 A}\).
Solution:

Question 11.
Show that \(\frac{1}{2 \sin 10^{\circ}}\) – 2 sin 70° = 1.
Solution:

Question 12.
Show that sin 19° + sin 41° + sin 83° = sin 23° + sin 37° + sin 79°.
Solution:

Question 13.
If sin A = \(\frac{1}{\sqrt{3}}\) and sin B = \(\frac{1}{\sqrt{5}}\) find the value of tan \(\frac { 1 }{ 2 }\)(A + B).cot\(\frac { 1 }{ 2 }\)(A – B).
Solution:

Question 14.
If sin θ = n sin (θ + 2α), show that (n – 1) tan (θ + α) + (n + 1) tan α = 0.
Solution:

Question 15.
If tan \(\frac { α }{ 2 }\) and tan\(\frac { ß }{ 2 }\) are the roots of the equation 8x² – 26x + 15 = 0, then find the value of cos (α + ß).
Solution:
Given tan\(\frac { α }{ 2 }\), tan\(\frac { ß }{ 2 }\) are the roots of quadratic eqn. 8x² – 26x +15 = 0

Question 16.
Prove that \(\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^n+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^n= \begin{cases}2 \cot ^n\left(\frac{A-B}{2}\right) & , \text { if } n \text { is even. } \\ 0 & \text {,if } n \text { is odd. }\end{cases}\)
Solution:

Question 17.
Find sin \(\frac { x }{ 2 }\), cos \(\frac { x }{ 2 }\) and tan \(\frac { x }{ 2 }\) in each of the following cases :
(i) sin x = \(\frac { 1 }{ 4 }\), x in II quadrant, (ii) tan x = \(\frac { -4 }{ 3 }\), x in II quadrant.
Solution:
(i) since x lies in IInd quadrant. ∴ \(\frac { x }{ 2 }\) lies in first quadrant.

(ii) Given tan x = – \(\frac { 4 }{ 3 }\)
since x lies in 2nd quadrant ∴ sin x > 0 and cos x < 0

Question 18.
Prove that cos 6x = 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1.
Solution:
cos 6x = cos (2 x 3x) = 2 cos² 3x – 1 = 2 [4 cos³ x – 3 cos x]² – 1
= 2 [16 cos⁶ x + 9 cos² x – 24 cos⁴ x] – 1
= 32 cos⁶ x + 18 cos² x – 48 cos⁴ x – 1

Question 19.
Prove that sin² 72° – sin² 60° = \(\frac{\sqrt{5}-1}{8}\).
Solution:

Question 20.
If tan x, tany, tan z are in GP., show that cos 2y = \(\frac{\cos (x+z)}{\cos (x-z)}\).
Solution:
Since tan x, tan y and tan z are in G.P

Question 21.
If tan \(\frac { α }{ 2 }\) : tan \(\frac { ß }{ 2 }\) = 1 : \(\sqrt{3}\) show that cos ß = \(\frac{2 \cos \alpha-1}{2-\cos \alpha}\).
Solution: