Students often turn to S Chand Class 11 Maths Solutions Chapter 5 Compound and Multiple Angles Ex 5(a) to clarify doubts and improve problem-solving skills.
S Chand Class 11 ICSE Maths Solutions Chapter 5 Compound and Multiple Angles Ex 5(a)
Question 1.
Compute :
(i) sin 15° from the functions of 60° and 45°,
(ii) cos 345° from the functions of 300° and 45°,
(iii) tan 105° from the functions of 45° and 60°,
(iv) sin 135° from the functions of 180° and 45°,
(v) cos 195° from the functions of 150° and 45°,
(vi) cosec (13π/12)
Solution:
(i) sin 15° = sin (60° – 45°) = sin 60° cos 45° – cos 60° sin 45°
= \(\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}-\frac{1}{2} \times \frac{1}{\sqrt{2}}=\frac{\sqrt{3}-1}{2 \sqrt{2}}\) [∵ sin (A – B) = sin A cos B – cos A sin B]
(ii) cos 345° = cos (300° + 45°) = cos 300° cos 45° – sin 300° sin 45°
= cos (270° + 30°) \(\frac{1}{\sqrt{2}}\) – sin (270° + 30°)\(\frac{1}{\sqrt{2}}\) = sin30° x \(\frac{1}{\sqrt{2}}\) + cos30° x \(\frac{1}{\sqrt{2}}\) [∵ cos (270° + θ) = cos θ and sin (270° + θ) = – cos θ]
= \(\frac{1}{2 \sqrt{2}}+\frac{\sqrt{3}}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
(iii) tan 105° = tan (60° + 45°) = \(\frac{\tan 60^{\circ}+\tan 45^{\circ}}{1-\tan 60^{\circ} \tan 45^{\circ}}=\frac{\sqrt{3}+1}{1-\sqrt{3}}\) [∵ tan (A + B) = \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) ]
= \(\frac{(\sqrt{3}+1)}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}}=\frac{3+1+2 \sqrt{3}}{1-3}=\frac{4+2 \sqrt{3}}{-2}=-(2+\sqrt{3})\)
(iv) sin 135° = sin (180°- 45°) = sin 180° cos 45° – cos 180° sin 45° = 0 x \(\frac{1}{\sqrt{2}}-(-1) \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\)
(v) cos 195° = cos (150° + 45°) = cos 150° cos 45° – sin 150° sin 45°
= cos (180°- 30°) cos 45° – sin (180°- 30°) sin 45°
= – cos 30° cos 45° – sin 30° sin 45°
= \(\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}-\frac{1}{2} \times \frac{1}{\sqrt{2}}=-\frac{(\sqrt{3}+1)}{2 \sqrt{2}}\)
Question 2.
Simplify by reducing to a single term :
(i) sin 3α cos 2α + cos 3α sin 2α
(ii) cos 5θ cos 2θ – sin 5θ sin 2θ.
(iii) sin 22° cos 38° + cos 22° sin 38°.
(iv) sin 80° cos 20° – cos 80° sin 20°.
(v) sin (x – y) cos x – cos (x -y) sin x.
(vi) cos (θ + α) cos (θ – α) – sin (θ + α) sin (θ – α)
(vii) \(\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}\)
(viii) \(\frac{\tan \alpha-\tan (\alpha-\beta)}{1+\tan \alpha \tan (\alpha-\beta)}\).
Solution:
(i) sin 3α cos 2α + cos 3α sin 2α = sin (3α + 2a) = sin 5a [∵ sin A cos B + cos A sin B = sin (A + B)]
(ii) cos 5θ cos 2θ – sin 5θ sin 2θ = cos (5θ + 2θ) = cos 7θ [∵ cos A cos B – sin A sin B = cos (A + B)]
(iii) sin 22° cos 38° + cos 22° sin 38° = sin (22° + 38°) = sin 60° = \(\frac{\sqrt{3}}{2}\)
(iv) sin 80° cos 20° – cos 80° sin 20° = sin (80° – 20°) = sin 60° = \(\frac{\sqrt{3}}{2}\)
(v) sin (x – y) cos x – cos (x – y) sin x = sin (x – y – x) = sin (- y) = – sin y [∵ sin A cos B – cos A sin B = sin (A – B)]
(vi) cos (θ + α) cos (θ – α) – sin (θ + α) sin (θ – α) = cos (θ + α + θ – α) = cos 2θ [∵ cos A cos B – sin A sin B = cos (A + B)]
(vii) \(\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}\) = tan (69° + 66°) [∵ tan (a + b) = \(\frac{\tan A+\tan B}{1-\tan A \tan B}\)]
= tan 135° = tan (180° – 45°) = – tan 45° = – 1
(viii) \(\frac{\tan \alpha-\tan (\alpha-\beta)}{1+\tan \alpha \tan (\alpha-\beta)}\) = tan [α – (α – ß)] = tan ß [∵ \(\frac{\tan A-\tan B}{1+\tan A \tan B}\) tan (A – B)]
Prove that:
Question 3.
(sin α cos ß + cos α sin ß) + (cos α cos ß – sin α sin ß)² = 1.
Solution:
L.H.S = (sin α cos ß + cos α sin ß)² + (cos α cos ß – sin α sin ß)²
= sin² (α + ß) + cos² (α + ß) = 1 = R.H.S.
Question 4.
sin (60° + θ) – sin (60° – θ) = sin θ.
Solution:
sin (60° + θ) – sin (60° – θ) = sin 60° cos θ + cos 60° sin θ – sin 60° cos θ + cos 60° sin θ
= 2 cos 60° sin θ = 2 x \(\frac { 1 }{ 2 }\) sin θ = sin θ
Question 5.
sin (θ + 30°) + cos (θ + 60°) = cos θ.
Solution:
L.H.S = sin (θ + 30°) + cos (θ + 60°)
= sin θ cos 30° + cos θ sin 30° + cos θ cos 60° – sin θ sin 60°
= sin θ x \(\frac{\sqrt{3}}{2}\) + \(\frac{\cos \theta}{2}\) + cos θ x \(\frac { 1 }{ 2 }\) – \(\frac{\sqrt{3}}{2}\) sin θ = cos θ = R.H.S.
Question 6.
sin (240° + θ) + cos (330° + θ) = 0.
Solution:
L.H.S. = sin (240° + θ) + cos (330° + θ) = sin (180° + 60° + θ) + cos (360° – 30° + θ)
= – sin (60° + θ) + cos [360° – (30° – θ)] = – sin (60° + θ) + cos (30° – θ) [∵ sin (180° + θ) = – sin θ ; cos (360° – θ) = cos θ]
= – sin 60° cos θ – cos 60° sin θ + cos 30° cos θ + sin θ sin 30°
= – \(\frac{\sqrt{3}}{2}\) cos θ – \(\frac { 1 }{ 2 }\) sin θ + \(\frac{\sqrt{3}}{2}\) cos θ + \(\frac { 1 }{ 2 }\) sin θ = 0 = R.H.S.
Question 7.
sin (A – 45°) = \(\frac{1}{\sqrt{2}}\) (sin A – cos A).
Solution:
sin (A – 45°) = sin A cos 45° – cos A sin 45° = \(\frac{\sin A}{\sqrt{2}}-\frac{\cos A}{\sqrt{2}}=\frac{1}{\sqrt{2}}\) (sin A – cos A)
Question 8.
\(\cos \left(\frac{\pi}{3}+x\right)=\frac{\cos x-\sqrt{3} \sin x}{2}\).
Solution:
Question 9.
(i) tan (45° + θ) = \(\frac{1+\tan \theta}{1-\tan \theta}\)
(ii) tan (45° – θ) = \(\frac{1-\tan \theta}{1+\tan \theta}\).
Solution:
Question 10.
(i) \(\frac{\sin (\theta+\phi)}{\sin \theta \cos \phi}\) = cot θ tan Φ + 1.
(ii) \(\frac{\sin (\theta-\phi)}{\sin \theta \sin \phi}\) = cot Φ – cot θ.
Solution:
(i) \(\frac{\sin (\theta+\phi)}{\sin \theta \cos \phi}=\frac{\sin \theta \cos \phi+\cos \theta \sin \phi}{\sin \theta \cos \phi}=\frac{\sin \theta \cos \phi}{\sin \theta \cos \phi}+\frac{\cos \theta \sin \phi}{\sin \theta \cos \phi}=1+\cot \theta \tan \phi\)
(ii) \(\frac{\sin (\theta-\phi)}{\sin \theta \sin \phi}=\frac{\sin \theta \cos \phi-\cos \theta \sin \phi}{\sin \theta \sin \phi}=\frac{\sin \theta \cos \phi}{\sin \theta \sin \phi}-\frac{\cos \theta \sin \phi}{\sin \theta \sin \phi}=\cot \phi-\cot \theta\)
Question 11.
\(\frac{\sin (A-B)}{\sin A \sin B}+\frac{\sin (B-C)}{\sin B \sin C}+\frac{\sin (C-A)}{\sin C \sin A}\) = 0.
Solution:
Question 12.
sin 105° + cos 105° = cos 45°.
Solution:
L.H.S. = sin 105° + cos 105° = sin (60° + 45°) + cos (60° + 45°)
= sin 60° cos 45° + cos 60° sin 45° + cos 60° cos 45° – sin 60° sin 45°
= \(\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}=\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}\) = cos 45° = R.H.S.
Question 13.
Find the value of sin (α + ß), cos (α + ß), and tan (α + ß), given
(i) sin α = \(\frac { 3 }{ 5 }\), cos ß = \(\frac { 5 }{ 13 }\), α and ß in Quadrant I.
(ii) cos α = \(\frac { -12 }{ 13 }\), cot ß = \(\frac { 24 }{ 7 }\), α in Quadrant II, ß in Quadrant III.
Solution:
(i) Given
(ii) since α lies in IInd quadrant
∴ sin α > 0
given cos α = \(\frac { -12 }{ 13 }\)
Question 14.
Find the values of sin (α – ß), cos (α – ß) and tan (α – ß), given
(i) sin α = \(\frac { 8 }{ 17 }\). tan ß = \(\frac { 5 }{ 12 }\), α and ß in Quadrant I.
(ii) cos α = \(\frac { -12 }{ 13 }\), cot ß = \(\frac { 24 }{ 7 }\), α in Quadrant II, ß in Quadrant I.
Solution:
(i) Since a lies in first quadrant ∴ cos α > 0
(ii) since a lies in 2nd quadrant ∴ sin α > 0
given cos α = – \(\frac { 12 }{ 13 }\)
Question 15.
If A and B are acute angles, find (A + B) given
(i) sin A = \(\frac{1}{\sqrt{5}}\), sin B = \(\frac{1}{\sqrt{10}}\).
(ii) tan A = \(\frac { 5 }{ 6 }\), tan B = \(\frac { 1 }{ 11 }\)
Solution:
Given A and B are acute angles
∴ cos A, cos B > 0
Question 16.
Given that tan α = \(\frac { m }{ m+1 }\), tan ß = \(\frac{1}{2 m+1}\) then is the value of α + ß?
Solution:
Question 17.
In the ∆ABC, the foot of the perpendicular from A to BC is D. Given that tan B = \(\frac { 4 }{ 3 }\), cos C = \(\frac { 15 }{ 17 }\) and that AB = 20 cm, calculate without using tables
(i) the lengths of the sides AC and BC ; (ii) the value of sin A.
Solution:
(i) Given tan ß = \(\frac { 4 }{ 3 }\) ; cos C = \(\frac { 15 }{ 17 }\) and AB = 20 cm
since tan ß = \(\frac { 4 }{ 3 }\) ⇒ \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{4}{3}\)
⇒ AD = 4k; BD = 3k
Since In ∆ADB, using pythagoras theorem,
AB² = AD² + BD²
⇒ 20² = (4k)² + (3k)²
⇒ 400 = 25k²
⇒ k² = \(\frac { 400 }{ 25 }\) = 16
⇒ k = 4
∴ AD = 16 ; BD = 12
Since cos C = \(\frac { 15 }{ 17 }\) = \(\frac { DC }{ AC }\)
⇒ DC = 15x ; AC = 17x
In ∆ADC, AC² = AD² + DC²
⇒ (17x)² = 16² + (15x)²
⇒ 289 x² – 225 x² = 256
⇒ 64x² = 256
⇒ x² = 4
⇒ x = 2
∴ AC = 17x = 17 x 2 = 34 and DC = 15 x 2 = 30
and BC = BD + DC = 12 + 30 = 42 cm
(ii) from figure ;
Question 18.
Given that tan (A + B) = 1 and tan (A – B) = \(\frac { 1 }{ 7 }\), find without using tables the values of tan A and tan B.
Solution:
Given tan (A + B) = 1 ⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = 1
⇒ tan A + tan B = 1 – tan A tan B … (1)
Also, tan (A – B) = \(\frac { 1 }{ 7 }\) ⇒ \(\frac{\tan A-\tan B}{1+\tan A \tan B}=\frac{1}{7}\)
⇒ 7 (tan A – tan B) = 1 + tan A tan B … (2)
On adding (1) and (2); we have
8 tan A – 6 tan B = 2 ⇒ 4 tan A – 3 tan B = 1 … (3)
From (1) and (2); we have
tan A + \(\frac{4 \tan A-1}{3}\) = 1 – tan A\(\left(\frac{4 \tan \mathrm{A}-1}{3}\right)\)
⇒ 3 tan A + 4 tan A – 1 = 3 – 4 tan² A + tan A
⇒ 4 tan² A + 6 tan A – 4 = 0
⇒ 2 tan² A + 3 tan A – 2 = 0
∴ tan A = \(\frac{-3 \pm \sqrt{9+16}}{4}=\frac{-3 \pm 5}{4}=\frac{1}{2},-2\)
When tan A = \(\frac { 1 }{ 2 }\) ∴ from (3); we have tan B = \(\frac{4 \tan A-1}{3}=\frac{2-1}{3}=\frac{1}{3}\)
When tan A = – 2 ∴ from (3); tan B = – 3
Question 19.
If tan (A + B) = x and tan B = \(\frac { 1 }{ 2 }\) prove that tan A = \(\frac{2 x-1}{x+2}\) and obtain an expression for tan (A – B) in terms of x. If tan (A – B) = \(\frac { 1 }{ 3 }\) and A is acute, find A without using tables.
Solution:
Question 20.
If sin (α + ß) = \(\frac { 4 }{ 5 }\), sin(α – ß) = \(\frac { 5 }{ 13 }\), α + ß, α – ß being acute angles prove that tan 2α = \(\frac { 63 }{ 16 }\).
Solution:
Given sin (α + ß) = \(\frac { 4 }{ 5 }\) and sin (α – ß) = \(\frac { 5 }{ 13 }\) since α + ß, α – ß are acute angles
Question 21.
Prove that 1 + tan 2θ tan P = sec 2θ.
Solution:
L.H.S = 1 + tan 2θ tan θ = 1 + \(\frac{\sin 2 \theta}{\cos 2 \theta} \cdot \frac{\sin \theta}{\cos \theta}=\frac{\cos 2 \theta \cos \theta+\sin 2 \theta \sin \theta}{\cos 2 \theta \cos \theta}\)
= \(\frac{\cos (2 \theta-\theta)}{\cos 2 \theta \cos \theta}=\frac{\cos \theta}{\cos 2 \theta \cos \theta}\) = sec 2θ = R.H.S
Question 22.
\(4 \sin \left(\frac{\pi}{3}-\theta\right) \sin \left(\frac{\pi}{3}+\theta\right)\) = 3 – 4 sin²θ.
Solution:
Question 23.
\(\frac{\cos 17^{\circ}+\sin 17^{\circ}}{\cos 17^{\circ}-\sin 17^{\circ}}\) = tan 62°.
Solution:
Question 24.
tan 3A – tan 2A – tan A = tan 3A tan 2 A tan A.
Solution:
tan 3A = tan (A + 2A) = \(\frac{\tan \mathrm{A}+\tan 2 \mathrm{~A}}{1-\tan \mathrm{A} \tan 2 \mathrm{~A}}\)
⇒ tan 3A [1 -tan A tan 2A] =tanA + tan2A
⇒ tan 3 A – tan A tan 2A tan 3A = tan A + tan 2A
⇒ tan 3 A – tan A – tan 2A = tan A tan 2A tan 3 A
Question 25.
tan 75° – tan 30° – tan 75° tan 30° = 1
Solution:
Question 26.
cos 2θ cos 2Φ + din² (θ – Φ) – sin² (θ + Φ) = cos (2θ + 2Φ).
Solution:
L.H.S = cos 2θ cos 2Φ + sin² (θ – Φ) – sin² (θ + Φ)
= cos 2θ cos 2Φ) + sin (θ – Φ + θ + Φ) sin (θ – Φ – θ – Φ) [∵ sin² A – sin² B = sin (A + B) sin (A – B)]
= cos 2θ cos 2Φ + sin 2θ sin (- 2Φ) = cos 2θ cos 2Φ – sin 2θ sin 2Φ
= cos (2θ + 2Φ) = R.H.S
Question 27.
If sin (θ + α) = cos (θ + α), prove that tan θ = \(\frac{1-\tan \alpha}{1+\tan \alpha}\).
Solution:
sin (θ + α) = cos (θ + α) ⇒ tan (θ + α) = 1 ⇒ θ + α = \(\frac { π }{ 4 }\) ⇒ θ = \(\frac { π }{ 4 }\) – α
⇒ tan θ = tan \(\tan \left(\frac{\pi}{4}-\alpha\right)=\frac{\tan \frac{\pi}{4}-\tan \alpha}{1+\tan \frac{\pi}{4} \tan \alpha} \Rightarrow \tan \theta=\frac{1-\tan \alpha}{1+\tan \alpha}\)
Question 28.
Given that A = B + C, prove that tan A – tan B – tan C = tan A tan B tan C.
Solution:
Given A = B + C ⇒ tan A = tan (B + C) = \(\frac{\tan B+\tan C}{1-\tan B \tan C}\)
⇒ tan A [1 – tan B tan C] = tan B + tan C
⇒ tan A – tan A tan B tan C = tan B + tan C
⇒ tan A – tan B – tan C = tan A tan B tan C
Question 29.
If A + B = 45°, show that tan A + tan B + tan A tan B = 1.
Hence, or otherwise, express tan 22°30′ in surd form.
Solution:
Given A + B = 45° + B) = tan 45° = 1 ⇒ ; — = 1
⇒ tan A + tan B = 1 – tan A tan B ⇒ tan A + tan B + tan A tan B = 1
putting A = B = 22° 30′ in eqn. (1); we have
2 tan 22° 30′ + tan2 22° 30′ = 1 ⇒ tan² θ + 2 tan θ – 1 = 0 ; where θ = 22° 30′
∴ tan θ = \(\frac{-2 \pm \sqrt{4+4}}{2}=\frac{-2 \pm 2 \sqrt{2}}{2}=-1 \pm \sqrt{2}\)
since 22° 30′ lies in first quadrant
∴ tan 22° 30′ > 0
Thus tan θ = tan (22° 30′) = \(\sqrt{2}\) – 1
Question 30.
(i) Use the expansion of tan (A – B) to find tan 15° without the use of tables, leaving your answer in surd form with an integral denominator.
(ii) Prove that \(\frac{2 \tan A}{1+\tan ^2 A}\) = sin 2A.
Solution:
Question 31.
If A + B = 225°, prove that tan A + tan B = 1 – tan A tan B.
Solution:
Given A + B = 225° ⇒ tan (A + B) = tan 225°
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = tan (180° + 45°) = tan 45° =1 ⇒ tan A + tan B = 1 – tan A tan B