OP Malhotra Class 11 Maths Solutions Chapter 5 Compound and Multiple Angles Ex 5(d)

Interactive S Chand Class 11 Maths Solutions Chapter 5 Compound and Multiple Angles Ex 5(d) engage students in active learning and exploration.

S Chand Class 11 ICSE Maths Solutions Chapter 5 Compound and Multiple Angles Ex 5(d)

Prove that

Question 1.
(i) (sin Φ – cos Φ)² = 1 – sin²Φ.
(ii) \(\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^2\) = 1 + sin θ.
Solution:
(i) (sin Φ – cos Φ))² = sin² Φ + cos² Φ – 2 sin Φ cos Φ = 1 – sin 2Φ
(ii) \(\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^2=\cos ^2 \frac{\theta}{2}+\sin ^2 \frac{\theta}{2}+2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}=1+\sin \left(2 \cdot \frac{\theta}{2}\right)=1+\sin \theta\)

Question 2.
\(\frac{\sin \theta \times \tan \frac{\theta}{2}}{2}=\sin ^2 \frac{\theta}{2}\)
Solution:
\(\frac{\sin \theta \tan \frac{\theta}{2}}{2}=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \tan \frac{\theta}{2}}{2}=\sin ^2 \frac{\theta}{2}\)

Question 3.
\(\frac{1}{\sec 2 \theta}=\cos ^4 \theta-\sin ^4 \theta\)
Solution:
\(\frac{1}{\sec 2 \theta}\) = cos 2θ = (cos² θ – sin² θ) . 1 = (cos² θ – sin² θ) (cos² θ + sin² θ) = cos⁴ θ – sin⁴ θ

Question 4.
\(\frac{2}{\cot \alpha \tan 2 \alpha}\) = 1 – tan² α.
Solution:
\(\frac{2}{\cot \alpha \tan 2 \alpha}=\frac{2 \tan \alpha}{\tan 2 \alpha}=\frac{2 \tan \alpha}{\frac{2 \tan \alpha}{1-\tan ^2 \alpha}}\) = 1 – tan² α

Question 5.
\(\frac{\sin 2 A}{1-\cos 2 A}\) = cot A
Solution:
\(\frac{\sin 2 A}{1-\cos 2 A}=\frac{2 \sin A \cos A}{2 \sin ^2 A}\) = cot A

Question 6.
\(\frac{\sin 2 A}{1+\cos 2 A}\) = tan A
Solution:
\(\frac{\sin 2 A}{1+\cos 2 A}=\frac{2 \sin A \cos A}{2 \cos ^2 A}\) = tan A

Question 7.
\(\frac{\sin 3 \theta}{\sin \theta}-\frac{\cos 3 \theta}{\cos \theta}\) = 2
Solution:

Question 8.
\(\frac{\cos ^3 A-\sin ^3 A}{\cos A-\sin A}=\frac{2+\sin 2 A}{2}\).
Solution:

Question 9.
\(\frac{1-\cos 2 \theta}{\sin 2 \theta}\) = tan θ.
Solution:
R.H.S = \(\frac{1-\cos 2 \theta}{\sin 2 \theta}=\frac{2 \sin ^2 \theta}{2 \sin \theta \cos \theta}\) = tan θ = R.H.S

Question 10.
cot α – tan α = 2 cot 2α.
Solution:
L.H.S = cot α – tan α = \(\frac{1}{\tan \alpha}-\tan \alpha=\frac{1-\tan ^2 \alpha}{\tan \alpha}=\frac{2\left(1-\tan ^2 \alpha\right)}{2 \tan \alpha}=\frac{2}{\frac{2 \tan \alpha}{1-\tan ^2 \alpha}}=\frac{2}{\tan 2 \alpha}\)
= 2 cot 2α = R.H.S

Question 11.
cosec A – 2 cot 2A cos A = 2 sin A.
Solution:
L.H.S = cosec A – 2 cot 2A cos A = \(\frac{1}{\sin A}-\frac{2 \cos 2 A \cos A}{\sin 2 A}=\frac{1}{\sin A}-\frac{2 \cos 2 A \cos A}{2 \sin A \cos A}\)
= \(\frac{1}{\sin A}-\frac{\cos 2 A}{\sin A}=\frac{1-\cos 2 A}{\sin A}=\frac{2 \sin ^2 A}{\sin A}\) = 2 sin A = R.H.S

Question 12.
\(\frac{1+\sin 2 x+\cos 2 x}{\cos x+\sin x}\) = 2 cos x.
Solution:
L.H.S = \(\frac{1+\sin 2 x+\cos 2 x}{\cos x+\sin x}=\frac{(1+\cos 2 x)+\sin 2 x}{\cos x+\sin x}=\frac{2 \cos ^2 x+2 \sin x \cos x}{\cos x+\sin x}\)
= \(\frac{2 \cos x(\cos x+\sin x)}{\cos x+\sin x}\) = 2 cos x = R.H.S

Question 13.
\(\frac{1-\cos 2 x+\sin x}{\sin 2 x+\cos x}\) = tan x.
Solution:
L.H.S = \(\frac{1-\cos 2 x+\sin x}{\sin 2 x+\cos x}=\frac{2 \sin ^2 x+\sin x}{2 \sin x \cos x+\cos x}=\frac{\sin x(2 \sin x+1)}{\cos x(2 \sin x+1)}\) = tan x = R.H.S

Question 14.
\(\frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}\) = tan θ
Solution:
L.H.S = \(\frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}=\frac{\sin \theta+2 \sin \theta \cos \theta}{1+\cos 2 \theta+\cos \theta}=\frac{\sin \theta(1+2 \cos \theta)}{2 \cos ^2 \theta+\cos \theta}=\frac{\sin \theta(1+2 \cos \theta)}{\cos \theta(1+2 \cos \theta)}\) = tan θ = R.H.S

Question 15.
cot A = \(\frac{1}{2}\left(\cot \frac{A}{2}-\tan \frac{A}{2}\right)\).
Solution:

Question 16.
\(\frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}\) = 3 tan 2A.
Solution:

Question 17.
\(\frac{\cos \mathbf{A}-\sin \mathbf{A}}{\cos \mathbf{A}+\sin \mathbf{A}}\) = sec 2A – tan 2A.
Solution:
R.H.S = sec 2A – tan 2A = \(\frac{1}{\cos 2 A}-\frac{\sin 2 A}{\cos 2 A}=\frac{1-\sin 2 A}{\cos 2 A}=\frac{\cos ^2 A+\sin ^2 A-2 \sin A \cos A}{\cos ^2 A-\sin ^2 A}\)
= \(\frac{(\cos A-\sin A)^2}{(\cos A-\sin A)(\cos A+\sin A)}=\frac{\cos A-\sin A}{\cos A+\sin A}\) = L.H.S

Question 18.
\(\frac{\sin 2 A}{1-\cos 2 A} \cdot \frac{1-\cos A}{\cos A}=\tan \frac{A}{2}\)
Solution:
L.H.S = \(\frac{\sin 2 A}{1-\cos 2 A} \cdot \frac{1-\cos A}{\cos A}=\frac{2 \sin A \cos A}{2 \sin ^2 A} \cdot \frac{2 \sin ^2 \frac{A}{2}}{\cos A}=\frac{2 \sin ^2 \frac{A}{2}}{2 \sin \frac{A}{2} \cos \frac{A}{2}}=\tan \frac{A}{2}\) = R.H.S

Question 19.
2 cos A = \(\sqrt{[2+\sqrt{2(1+\cos 4 A)}]}\), A ∈ I or IV Quad.
Solution:
R.H.S = \(\sqrt{2+\sqrt{2(1+\cos 4 \mathrm{~A})}}=\sqrt{2+\sqrt{2 \cdot 2 \cos ^2 2 \mathrm{~A}}}\)
= \(\sqrt{2+2 \cos 2 A}=\sqrt{2(1+\cos 2 A)}=\sqrt{2 \cdot 2 \cos ^2 A}\) = 2 cos A = L.H.S

Question 20.
tan 2A = (sec 2A + 1) \(\sqrt{\left(\sec ^2 A-1\right)}\), A in I or IV Quad.
Solution:
R.H.S = (sec 2A + 1) . \(\sqrt{\sec ^2 A-1}=\left(\frac{1}{\cos 2 A}+1\right) \tan A=\frac{1+\cos 2 A}{\cos 2 A} \cdot \tan A\)
= \(\frac{2 \cos ^2 A \cdot \tan A}{\cos ^2 A-\sin ^2 A}=\frac{2 \tan A \cos ^2 A}{\cos ^2 A\left[1-\tan ^2 A\right]}=\frac{2 \tan A}{1-\tan ^2 A}\) = tan 2A = L.H.S

Question 21.
\(\frac{\sec 8 A-1}{\sec 4 A-1}=\frac{\tan 8 A}{\tan 2 A}\).
Solution:

Question 22.
tan 2A – sec A sin A = tan A sec 2 A.
Solution:

Question 23.
(i) (cos A + cos B)² + (sin A + sin B)² = 4 cos² \(\left(\frac{\mathbf{A}-\mathbf{B}}{2}\right)\).
(ii) (cos A – cos B)² + (sin A – sin B)² = 4 sin² \(\left(\frac{\mathbf{A}-\mathbf{B}}{2}\right)\).
Solution:

Question 24.
(i) \(\cos ^2 \frac{\pi}{8}+\cos ^2 \frac{3 \pi}{8}+\cos ^2 \frac{5 \pi}{8}+\cos ^2 \frac{7 \pi}{8}\) = 2.
(ii) \(\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}=\frac{3}{2}\)
Solution:

Question 25.
cos α cos (60° – α) cos (60° + α) = \(\frac { 1 }{ 4 }\) cos 3α.
Solution:

Question 26.
\(\frac{\cos ^3 \alpha-\cos 3 \alpha}{\cos \alpha}+\frac{\sin ^3 \alpha+\sin 3 \alpha}{\sin \alpha}\) = 3.
Solution:

Question 27.
cos² 2θ – sin² θ = cos θ . cos 3θ.
Solution:
R.H.S = cos θ cos 3θ = cos θ (4 cos³ θ – 3 cos θ) = 4 cos⁴ θ – 3 cos² θ
L.H.S = cos² 2θ – sin² θ = (2 cos² θ – 1)² – (1 – cos² θ)
= 4 cos⁴ θ – 4 cos² θ + 1 – 1 + cos² θ
= 4 cos⁴ θ – 3 cos² θ
∴ L.H.S = R.H.S

Question 28.
1 + \(\frac{\cos 2 \theta+\cos 6 \theta}{\cos 4 \theta}=\frac{\sin 3 \theta}{\sin \theta}\).
Solution:

Question 29.
\(\cos ^3\left(x-\frac{2 \pi}{3}\right)+\cos ^3 x+\cos ^3\left(x+\frac{2 \pi}{3}\right)=\frac{3}{4} \cos 3 x\)
Solution:

Question 30.
\(\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}\) = 4.
Solution:

Question 31.
tan 70° – tan 20° – 2 tan 40° = 4 tan 10°.
Solution:

Question 32.
tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ = cot θ.
Solution:
We know that cot θ – tan θ = 2 cot 2θ … (1)
L.H.S = tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ
= cot θ – (cot θ – tan θ) + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ
= cot θ – 2 cot 2θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ [using (1)]
= cot θ – 2 [cot 2θ – tan 2θ] + 4 tan 4θ + 8 cot 8θ
= cot θ – 2 x 2 cot 4θ + 4 tan 4θ + 8 cot 8θ = cot θ – 4 [cot 4θ – tan 4θ] + 8 cot 8θ
= cot θ – 8 cot 8θ + 8 cot 8θ = cot θ = R.H.S

Question 33.
(i) 4 (cos³ 20° + cos³ 40°) = 3 (cos 20° + cos 40°).
(ii) 4 (cos³ 10° + sin³ 20°) = 3 (cos 10° + sin 20°).
Solution:
(i) We know that cos 3θ = 4 cos³ θ – 3 cos θ
L.H.S = 4 (cos³ 20° + cos³ 40°) = [cos (3 x 20°) + 3 cos 20°] + [cos (3 x 40°) + 3 cos 40°]
= [cos 60° + 3 cos 20°] + [cos 120° + 3 cos 40°]
= \(\frac { 1 }{ 2 }\) + 3 cos 20° + cos (180° – 60°) + 3 cos 40° = \(\frac { 1 }{ 2 }\) + 3 cos 20° – \(\frac { 1 }{ 2 }\) + 3 cos 40°
= 3 (cos 20° + cos 40°) = R.H.S

(ii) L.H.S. = 4 cos³ 10° + 4 sin3 20° = [cos (3 x 10°) + 3 cos 10°] + [3 sin 20° – sin 60°] [∵ cos 3θ = 4 cos³ θ – 3 cos θ ; sin 3θ = 3 sin θ – 4 sin³ θ]
= \(\frac{\sqrt{3}}{2}\) + 3 cos 10° + 3 sin 20° – \(\frac{\sqrt{3}}{2}\) = 3 (cos 10° + sin 20°) = R.H.S

Question 34.
cos³ x sin² x = \(\frac { 1 }{ 16 }\) (2 cos x – cos 3x – cos 5x).
Solution:

Question 35.
cot A + cot (60° + A) + cot (120° + A) = 3 cot 3A.
Solution:

Question 36.
(i) cos 36° – sin 18° = \(\frac { 1 }{ 2 }\).
(ii) cos² 36° + sin² 18° = \(\frac { 3 }{ 4 }\).
(iii) 3 cos 72° – 4 sin³ 18° = cos 36°.
(iv) cos² 48° – sin² 12° = \(\frac{\sqrt{5}+1}{8}\).
(v) sin \(\frac { 1 }{ 10 }\) π cos \(\frac { 1 }{ 5 }\) π = \(\frac { 1 }{ 4 }\).
(vi) sec 72° – sec 36° = 2.
(vii) sin² 24° – sin² 6° = \(\frac{\sqrt{5}-1}{8}\).
(viii) sin² \(\frac { π }{ 5 }\) sin²\(\frac { 2π }{ 5 }\) = \(\frac { 5 }{ 16 }\).
Solution:
(i) L.H.S = cos 36° – sin 18° = \(\frac{\sqrt{5}+1}{4}-\frac{\sqrt{5}-1}{4}=\frac{\sqrt{5}+1-\sqrt{5}+1}{4}=\frac{1}{2}\) = R.H.S

(ii) L.H.S. = cos² 36° + sin² 18° = \(\left(\frac{\sqrt{5}+1}{4}\right)^2+\left(\frac{\sqrt{5}-1}{4}\right)^2\)
= \(\frac{1}{16}[5+1+2 \sqrt{5}+5+1-2 \sqrt{5}]=\frac{12}{16}=\frac{3}{4}\)

(iii) L.H.S = 3 cos 72° – 4 sin3 18° = 3 cos (90° – 18°) – 4 sin3 18° = 3 sin 18° – 4 sin3 18°
= sin (3 x 18°) [∵ 3 sin θ – 4 sin³ θ = sin 3θ]
= sin 54° = sin (90° – 36°) = cos 36° = R.H.S

(iv) L.H.S = cos2 48° – sin2 12° = cos (48° + 12°) cos (48° – 12°) [∵ cos (A + B) cos (A – B) = cos² A – sin² B]
= cos 60° cos 36° = \(\frac{1}{2} \times \frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}+1}{8}\) = R.H.S

Question 37.
(i) sin 12° sin 48° sin 54° = \(\frac { 1 }{ 8 }\).
(ii) 4 cos 6° cos 42° cos 66° cos 78° = \(\frac { 1 }{ 4 }\).
(iii) \(\sin \frac{\pi}{5} \sin \frac{2 \pi}{5} \sin \frac{4 \pi}{5} \sin \frac{3 \pi}{5}=\frac{5}{16}\)
(iv) \(\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}=\frac{1}{16}\)
(v) \(\cos ^2 \frac{\pi}{10}+\cos ^2 \frac{2 \pi}{5}+\cos ^2 \frac{3 \pi}{5}+\cos ^2 \frac{9 \pi}{10}\) = 2
(vi) tan 6° tan 42° tan 66° tan 78° = 1
Solution:

Question 38.
cos α cos 2α cos 4α cos 8α = \(\frac { 1 }{ 16 }\), if a = 24°.
Solution:

Question 39.
cos 12° cos 24° cos 36° cos 48° cos 72° cos 84° = \(\frac{1}{2^6}\)
Solution:

Question 40.
If θ = \(\frac{\pi}{2^n+1}\) prove that 2ⁿ cos θ cos 2θ cos 2² θ ………. cos 2^n-1 θ = 1.
Solution:

Question 41.
If tan \(\frac{x-y}{2}\), tan z, tan \(\frac{x+y}{2}\) are in GP., then show that cos x = cos.y. cos 2z.
Solution:

Question 42.
If sin θ is GM. of sin Φ and cos Φ, then prove that cos 2θ = 2 cos²(\(\frac { π }{ 4 }\) + Φ)
Solution:

Question 43.
If m tan (θ – 30°) = n tan (θ + 120°), show that cos 2θ = \(\frac{m+n}{2(m-n)}\).
Solution:

Question 44.
If sin α = λ sin (θ – α) then prove that \(\tan \left(\alpha-\frac{\theta}{2}\right)=\frac{\lambda-1}{\lambda+1} \tan \frac{\theta}{2}\).
Solution:

Question 45.
If tan \(\frac{\alpha}{2}=\sqrt{\left\{\left(\frac{1-e}{1+e}\right)\right\}}\) tan \(\frac { ß }{ 2 }\), Prove that cos ß = \(\frac{\cos \alpha-e}{1-e \cos \alpha}\).
Solution: