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Class 11 RD Sharma Solutions – Chapter 8 Transformation Formulae – Exercise 8.2 | Set 1

  • Last Updated : 19 Jul, 2021

Question 1. Express each of the following as the product of sines and cosines:

(i) sin 12θ + sin 4θ

Solution:

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 12θ + sin 4θ = 2 sin (12θ + 4θ)/2 cos (12θ – 4θ)/2

= 2 sin 16θ/2 cos 8θ/2

= 2 sin 8θ cos 4θ



(ii) sin 5θ – sin θ

Solution:

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

sin 5θ – sin θ = 2 cos (5θ + θ)/2 sin (5θ – θ)/2

= 2 cos 6θ/2 sin 4θ/2

= 2 cos 3θ sin 2θ

(iii) cos 12θ + cos 8θ

Solution:

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 12θ + cos 8θ = 2 cos (12θ + 8θ)/2 cos (12θ – 8θ)/2



= 2 cos 20θ/2 cos 4θ/2

= 2 cos 10θ cos 2θ

(iv) cos 12θ – cos 4θ

Solution:

We know, cos A – cos B = –2 sin (A+B)/2 sin (A–B)/2

cos 12θ – cos 4θ = –2 sin (12θ + 4θ)/2 sin (12θ – 4θ)/2

= –2 sin 16θ/2 sin 8θ/2

= –2 sin 8θ sin 4θ

(v) sin 2θ + cos 4θ

Solution:

sin 2θ + cos 4θ = sin 2θ + sin (90o – 4θ)

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 2θ + sin (90o – 4θ) = 2 sin (2x + 90o – 4θ)/2 cos (2θ – 90o + 4θ)/2

= 2 sin (90o – 2θ)/2 cos (6θ – 90o)/2

= 2 sin (45o – θ) cos (3θ – 45o)

= 2 sin (45o – θ) cos [–(45o – 3θ)]

= 2 sin (45o – θ) cos (45o – 3θ)

= 2 sin (π/4 – θ) cos (π/4 – 3θ)

Question 2. Prove that :

(i) sin 38° + sin 22° = sin 82°

Solution:

Given, L.H.S. = sin 38° + sin 22°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 38° + sin 22° = 2 sin (38o + 22o)/2 cos (38o – 22o)/2



= 2 sin 60o/2 cos 16o/2

= 2 sin 30o cos 8o

= 2 × (1/2) × cos 8o

= cos 8o

= cos (90° – 82°)

= sin 82°

= R.H.S.

Hence proved.

(ii) cos 100° + cos 20° = cos 40°

Solution:

Given, L.H.S. = cos 100° + cos 20°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 100° + cos 20° = 2 cos (100o + 20o)/2 cos (100o – 20o)/2

= 2 cos 120o/2 cos 80o/2

= 2 cos 60o cos 40o

= 2 × (1/2) × cos 40o

= cos 40o

= R.H.S.

Hence Proved.

(iii) sin 50° + sin 10° = cos 20°

Solution:

Given, L.H.S. = sin 50° + sin 10°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2.

sin 50° + sin 10° = 2 sin (50o + 10o)/2 cos (50o – 10o)/2

= 2 sin 60o/2 cos 40o/2

= 2 sin 30o cos 20o

= 2 × (1/2) × cos 20o

= cos 20o

= R.H.S

Hence Proved.

(iv) sin 23° + sin 37° = cos 7°

Solution:

Given L.H.S. = sin 23° + sin 37°.



We know sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 23° + sin 37° = 2 sin (23o + 37o)/2 cos (23o – 37o)/2

= 2 sin 60o/2 cos (–14o/2)

= 2 sin 30o cos (–7o)

= 2 × (1/2) × cos 7o

= cos 7o

= R.H.S.

Hence Proved.

(v) sin 105° + cos 105° = cos 45°

Solution:

Given, L.H.S. = sin 105° + cos 105°

sin 105° + cos 105° = sin 105o + sin (90o – 105o)

= sin 105o + sin (–15o)

= sin 105o – sin 15o

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

sin 105o – sin 15o = 2 cos (105o + 15o)/2 sin (105o – 15o)/2

= 2 cos 120o/2 sin 90o/2

= 2 cos 60o sin 45o

= 2 × (1/2) × (1/√2)

= 1/√2

= cos 45o

= R.H.S.

Hence Proved.

(vi) sin 40° + sin 20° = cos 10°

Solution:

Given, L.H.S. = sin 40° + sin 20°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 40° + sin 20° = 2 sin (40o + 20o)/2 cos (40o – 20o)/2

= 2 sin 60o/2 cos 20o/2

= 2 sin 30o cos 10o

= 2 × (1/2) × cos 10o

= cos 10o

= R.H.S.

Hence Proved.

Question 3. Prove that:

(i) cos 55° + cos 65° + cos 175° = 0

Solution:

Given, L.H.S. = cos 55° + cos 65° + cos 175°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 55° + cos 65° + cos 175° = 2 cos (55o + 65o)/2 cos (55o – 65o) + cos (180o – 5o)

= 2 cos 120o/2 cos (–10o)/2 – cos 5o

= 2 cos 60° cos (–5°) – cos 5°

= 2 × (1/2) × cos 5o – cos 5o

= cos 5o – cos 5o



= 0

= R.H.S.

Hence Proved.

(ii) sin 50° – sin 70° + sin 10° = 0

Solution:

Given, L.H.S. = sin 50° – sin 70° + sin 10°.

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

sin 50° – sin 70° + sin 10° = 2 cos (50o + 70o)/2 sin (50o – 70o) + sin 10o

= 2 cos 120o/2 sin (–20o)/2 + sin 10o

= 2 cos 60o (–sin 10o) + sin 10o

= 2 × (1/2) × (–sin 10o) + sin 10o

= 0

= R.H.S.

Hence Proved.

(iii) cos 80° + cos 40° – cos 20° = 0

Solution:

Given L.H.S. = cos 80° + cos 40° – cos 20°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 80° + cos 40° – cos 20° = 2 cos (80o + 40o)/2 cos (80o – 40o) – cos 20o

= 2 cos 120o/2 cos 40o/2 – cos 20o

= 2 cos 60° cos 20o – cos 20°

= 2 × (1/2) × cos 20o – cos 20o

= 0

= R.H.S.

Hence Proved.

(iv) cos 20° + cos 100° + cos 140° = 0

Solution:

Given, L.H.S. = cos 20° + cos 100° + cos 140°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2.

cos 20° + cos 100° + cos 140° = 2 cos (20o + 100o)/2 cos (20o – 100o) + cos (80o – 40o)

= 2 cos 120o/2 cos (–80o)/2 – cos 40o

= 2 cos 60° cos (–40°) – cos 40°

= 2 × (1/2) × cos 40o – cos 40o

= 0

= R.H.S.

Hence Proved.

(v) sin 5π/18 – cos 4π/9 = √3 sin π/9

Solution:

Given, L.H.S. = sin 5π/18 – cos 4π/9

= sin 5π/18 – sin (π/2 – 4π/9)

= sin 5π/18 – sin (9π – 8π)/18

= sin 5π/18 – sin π/18

We know, sin A – sin B = 2 cos (A+B)/2 sin (A– B)/2

= 2 cos (6π/36) sin (4π/36)



= 2 cos π/6 sin π/9

= 2 cos 30o sin π/9

= 2 × (√3/2) × sin π/9

= √3 sin π/9

= R.H.S.

Hence Proved.

(vi) cos π/12 – sin π/12 = 1/√2

Solution:

Given, cos π/12 – sin π/12 = sin (π/2 – π/12) – sin π/12

= sin (6π – 5π)/12 – sin π/12

= sin 5π/12 – sin π/12

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

= 2 cos (6π/24) sin (4π/24)

= 2 cos π/4 sin π/6

= 2 cos 45o sin 30o

= 2 × (1/√2) × (1/2)

= 1/√2

= R.H.S.

Hence Proved.

(vii) sin 80° – cos 70° = cos 50°

Solution:

We have, sin 80° = cos 50° + cos 70o

Here, R.H.S. = cos 50° + cos 70o

We know,

cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 50° + cos 70o = 2 cos (50o + 70o)/2 cos (50o – 70o)/2

= 2 cos 120o/2 cos (–20o)/2

= 2 cos 60o cos (–10o)

= 2 × (1/2) × cos 10o

= cos 10o

= cos (90° – 80°)

= sin 80°

= L.H.S.

Hence Proved.

(viii) sin 51° + cos 81° = cos 21°

Solution:

Given, L.H.S. = sin 51° + cos 81°

= sin 51o + sin (90o – 81o)

= sin 51o + sin 9o

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 51o + sin 9o = 2 sin (51o + 9o)/2 cos (51o – 9o)/2

= 2 sin 60o/2 cos 42o/2

= 2 sin 30o cos 21o

= 2 × (1/2) × cos 21o

= cos 21o

= R.H.S.

Hence Proved.

Question 4. Prove that:

(i) cos (3π/4 + x) – cos (3π/4 – x) = –√2 sin x

Solution:

Given, L.H.S. = cos (3π/4 + x) – cos (3π/4 – x)

We know, cos A – cos B = –2 sin (A+B)/2 sin (A–B)/2

cos (3π/4 + x) – cos (3π/4 – x) = –2 sin (3π/4 + x + 3π/4 – x)/2 sin (3π/4 + x – 3π/4 + x)/2

= –2 sin (6π/4)/2 sin 2x/2

= –2 sin 6π/8 sin x

= –2 sin 3π/4 sin x

= –2 sin (π – π/4) sin x

= –2 sin π/4 sin x

= –2 × (1/√2) × sin x

= –√2 sin x

= R.H.S.

Hence proved.

(ii) cos (π/4 + x) + cos (π/4 – x) = √2 cos x

Solution:

Given, L.H.S. = cos (π/4 + x) + cos (π/4 – x)

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2



cos (π/4 + x) + cos (π/4 – x) = 2 cos (π/4 + x + π/4 – x)/2 cos (π/4 + x – π/4 + x)/2

= 2 cos (2π/4)/2 cos 2x/2

= 2 cos 2π/8 cos x

= 2 sin π/4 cos x

= 2 × (1/√2) × cos x

= √2 cos x

= R.H.S.

Hence proved.

Question 5. Prove that:

(i) sin 65o + cos 65o = √2 cos 20o

Solution:

Given L.H.S. = sin 65o + cos 65o

= sin 65o + sin (90o – 65o)

= sin 65o + sin 25o

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 65o + sin 25o = 2 sin (65o + 25o)/2 cos (65o – 25o)/2

= 2 sin 90o/2 cos 40o/2

= 2 sin 45o cos 20o

= 2 × (1/√2) × cos 20o

= √2 cos 20o

= R.H.S.

Hence proved.

(ii) sin 47o + cos 77o = cos 17o

Solution:

Given, L.H.S. = sin 47o + cos 77o

= sin 47o + sin (90o – 77o)

= sin 47o + sin 13o

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 47o + sin 13o = 2 sin (47o + 13o)/2 cos (47o – 13o)/2

= 2 sin 60o/2 cos 34o/2

= 2 sin 30o cos 17o

= 2 × (1/2) × cos 17o

= cos 17o

= R.H.S.

Hence proved.

Question 6. Prove that:

(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A

Solution:

Given, L.H.S. = cos 3A + cos 5A + cos 7A + cos 15A

= (cos 5A + cos 3A) + (cos 15A + cos 7A)

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

= (cos 5A + cos 3A) + (cos 15A + cos 7A)

= [2 cos (5A+3A)/2 cos (5A–3A)/2] + [2 cos (15A+7A)/2 cos (15A–7A)/2]

= [2 cos 8A/2 cos 2A/2] + [2 cos 22A/2 cos 8A/2]

= [2 cos 4A cos A] + [2 cos 11A cos 4A]

= 2 cos 4A (cos 11A + cos A)

= 2 cos 4A [2 cos (11A+A)/2 cos (11A-A)/2]

= 2 cos 4A [2 cos 12A/2 cos 10A/2]

= 2 cos 4A [2 cos 6A cos 5A]

= 4 cos 4A cos 5A cos 6A

= R.H.S.

Hence proved.

(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A

Solution:

Given L.H.S. = cos A + cos 3A + cos 5A + cos 7A

= (cos 3A + cos A) + (cos 7A + cos 5A)

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

= (cos 3A + cos A) + (cos 7A + cos 5A)

= [2 cos (3A+A)/2 cos (3A–A)/2] + [2 cos (7A+5A)/2 cos (7A–5A)/2]

= [2 cos 4A/2 cos 2A/2] + [2 cos 12A/2 cos 2A/2]

= [2 cos 2A cos A] + [2 cos 6A cos A]

= 2 cos A (cos 6A + cos 2A)

= 2 cos A [2 cos (6A+2A)/2 cos (6A–2A)/2]

= 2 cos A [2 cos 8A/2 cos 4A/2]

= 2 cos A [2 cos 4A cos 2A]

= 4 cos A cos 2A cos 4A



= R.H.S.

Hence proved.

(iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos A/2 cos 3A/2 sin 3A

Solution:

Given, L.H.S. = sin A + sin 2A + sin 4A + sin 5A

= (sin 2A + sin A) + (sin 5A + sin 4A)

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

= (sin 2A + sin A) + (sin 5A + sin 4A)

= [2 sin (2A+A)/2 cos (2A–A)/2] + [2 sin (5A+4A)/2 cos (5A–4A)/2]

= [2 sin 3A/2 cos A/2] + [2 sin 9A/2 cos A/2]

= 2 cos A/2 (sin 9A/2 + sin 3A/2)

= 2 cos A/2 [2 sin (9A/2 + 3A/2)/2 cos (9A/2 – 3A/2)/2]

= 2 cos A/2 [2 sin ((9A+3A)/2)/2 cos ((9A–3A)/2)/2]

= 2 cos A/2 [2 sin 12A/4 cos 6A/4]

= 2 cos A/2 [2 sin 3A cos 3A/2]

= 4 cos A/2 cos 3A/2 sin 3A

= R.H.S.

Hence proved.

(iv) sin 3A + sin 2A – sin A = 4 sin A cos A/2 cos 3A/2

Solution:

Given, L.H.S. = sin 3A + sin 2A – sin A

= (sin 3A – sin A) + sin 2A

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

= (sin 3A – sin A) + sin 2A

= 2 cos (3A + A)/2 sin (3A – A)/2 + sin 2A

= 2 cos 4A/2 sin 2A/2 + sin 2A

= 2 cos 2A sin A + 2 sin A cos A

= 2 sin A (cos 2A + cos A)

= 2 sin A [2 cos (2A+A)/2 cos (2A-A)/2]

= 2 sin A [2 cos 3A/2 cos A/2]

= 4 sin A cos A/2 cos 3A/2

= R.H.S.

Hence proved.

(v) cos 20o cos 100o + cos 100o cos 140o – cos 140o cos 200o = – 3/4

Solution:

Given L.H.S. = cos 20o cos 100o + cos 100o cos 140o – cos 140o cos 200o

= 1/2 [2 cos 100o cos 20o + 2 cos 140o cos 100o – 2 cos 200o cos 140o]

We know that, 2 cos A cos B = cos (A+B) + cos (A–B)

= 1/2 [cos (100o + 20o) + cos (100o – 20o) + cos (140o + 100o) + cos (140o – 100o) – cos (200o + 140o) – cos (200o – 140o)]]

= 1/2 [cos 120o + cos 80o + cos 240o + cos 40o – cos 340o – cos 60o]

= 1/2 [cos (90o + 30o) + cos 80o + cos (180o + 60o) + cos 40o – cos (360o – 20o) – cos 60o]

= 1/2 [–sin 30o + cos 80o – cos 60o + cos 40o – cos 20o – cos 60o]

= 1/2 [–sin 30o + cos 80o + cos 40o – cos 20o – 2 cos 60o]

= 1/2 [–sin 30o + 2 cos (80o+40o)/2 cos (80o–40o)/2 – cos 20o – 2 × 1/2]

= 1/2 [–sin 30o + 2 cos 120o/2 cos 40o/2 – cos 20o – 1]

= 1/2 [–sin 30o + 2 cos 60o cos 20o – cos 20o – 1]

= 1/2 [–1/2 + 2×(1/2)×cos 20o – cos 20o – 1]

= 1/2 [–1/2 + cos 20o – cos 20o – 1]

= 1/2 [–1/2 –1]

= 1/2 [–3/2]

= –3/4

= R.H.S.

Hence proved.

(vi) sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2 = sin 2x sin 5x

Solution:

Given L.H.S. = sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2

= 1/2 [2 sin 7x/2 sin x/2 + 2 sin 11x/2 sin 3x/2]

We know, 2 sin A sin B = cos (A–B) – cos (A+B)

= 1/2 [cos (7x/2 – x/2) – cos (7x/2 + x/2) + cos (11x/2 – 3x/2) – cos (11x/2 + 3x/2)]

= 1/2 [cos (7x–x)/2 – cos (7x+x)/2 + cos (11x–3x)/2 – cos (11x+3x)/2]

= 1/2 [cos 6x/2 – cos 8x/2 + cos 8x/2 – cos 14x/2]

= 1/2 [cos 3x – cos 7x]

= –1/2 [cos 7x – cos 3x]

= –1/2 [–2 sin (7x+3x)/2 sin (7x–3x)/2]



= –1/2 [–2 sin 10x/2 sin 4x/2]

= –1/2 [–2 sin 5x sin 2x]

= –2/–2 sin 5x sin 2x

= sin 2x sin 5x

= R.H.S.

Hence proved.

(vii) cos x cos x/2 – cos 3x cos 9x/2 = sin 4x sin 7x/2

Solution:

Given L.H.S. = cos x cos x/2 – cos 3x cos 9x/2

= 1/2 [2 cos x cos x/2 – 2 cos 9x/2 cos 3x]

We know, 2 cos A cos B = cos (A+B) + cos (A–B)

= 1/2 [cos (x + x/2) + cos (x – x/2) – cos (9x/2 + 3x) – cos (9x/2 – 3x)]

= 1/2 [cos (2x+x)/2 + cos (2x–x)/2 – cos (9x+6x)/2 – cos (9x–6x)/2]

= 1/2 [cos 3x/2 + cos x/2 – cos 15x/2 – cos 3x/2]

= 1/2 [cos x/2 – cos 15x/2]

= – 1/2 [cos 15x/2 – cos x/2]

= – 1/2 [–2 sin (15x/2 + x/2)/2 sin (15x/2 – x/2)/2]

= -1/2 [–2 sin (16x/2)/2 sin (14x/2)/2]

= -1/2 [–2 sin 16x/4 sin 7x/2]

= – 1/2 [–2 sin 4x sin 7x/2]

= –2/–2 [sin 4x sin 7x/2]

= sin 4x sin 7x/2

= R.H.S.

Hence proved.

Question 7. Prove that:

(i) \frac{\sin A + \sin 3A}{\cos A - \cos 3A} = \cot A

Solution:

We have,

L.H.S. = \frac{\sin A + \sin 3A}{\cos A - \cos 3A}

\frac{2\sin \left( \frac{A + 3A}{2} \right) \cos \left( \frac{A - 3A}{2} \right)}{2\sin \left( \frac{A + 3A}{2} \right) \sin \left( \frac{3A - A}{2} \right)}

\frac{\sin 2A \cos \left( - A \right)}{\sin 2A \sin A}

\frac{\sin 2A \cos A}{\sin 2A \sin A}

= cot A



= R.H.S.

Hence proved.

(ii) \frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A} = \cot 8A

Solution:

We have,

L.H.S. = \frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A}

\frac{2\sin \left( \frac{9A - 7A}{2} \right) \cos \left( \frac{9A + 7A}{2} \right)}{2\sin \left( \frac{7A + 9A}{2} \right) \sin \left( \frac{9A - 7A}{2} \right)}

\frac{\sin A \cos 8A}{\sin 8A \sin A}

= cot 8A

= R.H.S.

Hence proved.

(iii)\frac{sinA-sinB}{cosA+cosB}=tan\frac{A-B}{2}

Solution:

We have,

L.H.S. =\frac{sinA-sinB}{cosA+cosB}

=\frac{2cos\frac{A+B}{2}sin\frac{A-B}{2}}{2cos\frac{A+B}{2}cos\frac{A-B}{2}}

=\frac{sin\frac{A-B}{2}}{cos\frac{A-B}{2}}

=tan\frac{A-B}{2}

= R.H.S.

Hence proved.

(iv)\frac{sinA+sinB}{sinA-sinB}=tan(\frac{A+B}{2})cot(\frac{A-B}{2})

Solution:

We have,



L.H.S. =\frac{sinA+sinB}{sinA-sinB}

=\frac{2sin\frac{A+B}{2}cos\frac{A-B}{2}}{2cos\frac{A+B}{2}sin\frac{A-B}{2}}

=\frac{sin\frac{A+B}{2}cos\frac{A-B}{2}}{cos\frac{A+B}{2}sin\frac{A-B}{2}}

=tan(\frac{A+B}{2})cot(\frac{A-B}{2})

= R.H.S.

Hence proved.

(iv)\frac{cosA+cosB}{cosB-cosA}=cot(\frac{A+B}{2})cot(\frac{A-B}{2})

Solution:

We have,

L.H.S. =\frac{cosA+cosB}{cosB-cosA}

=\frac{2cos\frac{A+B}{2}cos\frac{A-B}{2}}{-2sin\frac{A+B}{2}sin\frac{B-A}{2}}

=\frac{cos\frac{A+B}{2}cos\frac{A-B}{2}}{sin\frac{A+B}{2}sin\frac{A-B}{2}}

=cot(\frac{A+B}{2})cot(\frac{A-B}{2})

= R.H.S.

Hence proved.

Question 8. Prove that:

(i) \frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A} = \tan 3A

Solution:

We have,

L.H.S. = \frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A}

\frac{\sin A + \sin 5A + \sin 3A}{\cos A + \cos 5A + \cos 3A}

\frac{2\sin \left( \frac{A + 5A}{2} \right) \cos \left( \frac{A - 5A}{2} \right) + \sin 3A}{2\cos \left( \frac{A + 5A}{2} \right) \cos \left( \frac{A - 5A}{2} \right) + \cos 3A}

\frac{2 \sin 3A \cos \left( - 2A \right) + \sin 3A}{2\cos 3A \cos \left( - 2A \right) + \cos 3A}



\frac{2\sin 3A \cos 2A + \sin 3A}{2\cos 3A \cos 2A + \cos 3A}

\frac{\sin 3A \left[ 2\cos 2A + 1 \right]}{\cos 3A \left[ 2\cos 2A + 1 \right]}

= tan 3A

= R.H.S.

Hence proved.

(ii)\frac{cos3A+2cos5A+cos7A}{cosA+2cos3A+cos5A}=\frac{cos5A}{cos3A}

Solution:

We have,

L.H.S. =\frac{cos3A+2cos5A+cos7A}{cosA+2cos3A+cos5A}

=\frac{2cos\frac{10A}{2}cos\frac{4A}{2}+2cos5A}{cos\frac{6A}{2}cos\frac{4A}{2}+2cos3A}

=\frac{2cos5Acos2A+2cos5A}{2cos3Acos2A+2cos3A}

=\frac{2cos5A(cos2A+1)}{2cos3A(cos2A+1)}

=\frac{cos5A}{cos3A}

= R.H.S.

Hence proved.

(iii) \frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A} = \cot 3A

Solution:

We have,

L.H.S. = \frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A}

\frac{\cos 4A + \cos 2A + \cos 3A}{\sin 4A + \sin 2A + \sin 3A}

\frac{2\cos \left( \frac{4A + 2A}{2} \right) \cos \left( \frac{4A - 2A}{2} \right) + \cos 3A}{2\sin \left( \frac{4A + 2A}{2} \right) \cos \left( \frac{4A - 2A}{2} \right) + \sin 3A}

\frac{2\cos 3A \cos A + \cos 3A}{2\sin 3A \cos A + \sin 3A}



\frac{\cos 3A\left[ 2\cos A + 1 \right]}{\sin 3A\left[ 2\cos A + 1 \right]}

= cot 3A

= R.H.S.

Hence proved.

(iv) \frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A} = \tan 6A

Solution:

We have,

L.H.S. = \frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A}

\frac{\sin 3A + \sin 9A + \sin 5A + \sin 7A}{\cos 3A + \cos 9A + \cos 5A + \sin 7A}

\frac{2\sin \left( \frac{3A + 9A}{2} \right) \cos \left( \frac{3A - 9A}{2} \right) + 2\sin \left( \frac{5A + 7A}{2} \right) \cos \left( \frac{5A - 7A}{2} \right)}{2\cos \left( \frac{3A + 9A}{2} \right) \cos \left( \frac{3A - 9A}{2} \right) + 2\cos \left( \frac{5A + 7A}{2} \right) \cos \left( \frac{5A - 7A}{2} \right)}

\frac{2\sin 6A \cos \left( - 3A \right) + 2\sin 6A \cos \left( - A \right)}{2\cos 6A \cos \left( - 3A \right) + 2\cos 6A \cos \left( - A \right)}



\frac{2\sin 6A \cos 3A + 2\sin 6A \cos A}{2\cos 6A \cos 3A + 2\cos 6A \cos A}

\frac{2\sin 6A\left[ \cos 3A + \cos A \right]}{2\cos 6A\left[ \cos 3A + \cos A \right]}

= tan 6A

= R.H.S.

Hence proved.

(v) \frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A + \cos 7A - \cos 5A - \cos 8A} = \cot 6A

Solution:

We have,

L.H.S. = \frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A + \cos 7A - \cos 5A - \cos 8A}

\frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A - \cos 8A + \cos 7A - \sin 5A}

\frac{2\sin \left( \frac{5A - 7A}{2} \right) \cos \left( \frac{5A + 7A}{2} \right) + 2\sin \left( \frac{8A - 4A}{2} \right) \cos \left( \frac{8A + 4A}{2} \right)}{- 2\sin \left( \frac{4A + 8A}{2} \right) \sin \left( \frac{4A - 8A}{2} \right) - 2\sin \left( \frac{7A + 5A}{2} \right) \sin \left( \frac{7A - 5A}{2} \right)}

\frac{2\sin \left( - A \right) \cos 6A + 2\sin 2A \cos 6A}{- 2\sin 6A \sin \left( - 2A \right) - 2\sin 6A \sin A}

\frac{- 2\sin A \cos 6A + 2\sin 2A cos 6A}{2\sin 6A \sin 2A - 2\sin 6A \sin A}

\frac{2\cos 6A\left[ \sin 2A - \sin A \right]}{2\sin 6A\left[ \sin 2A - \sin A \right]}

= cot 6A

= R.H.S.

Hence proved.

(vi) \frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A} = \tan A

Solution:

We have,

L.H.S. = \frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A}

Multiplying numerator and denominator by 2, we get



\frac{2\sin 5A \cos 2A - 2\sin 6A \cos A}{2\sin A \sin 2A - 2\cos 2A \cos 3A}

\frac{\sin \left( 5A + 2A \right) + \sin \left( 5A - 2A \right) - \sin \left( 6A + A \right) - \sin \left( 6A - A \right)}{\cos \left( A - 2A \right) + \cos \left( A + 2A \right) - \cos \left( 2A + 3A \right) - \cos \left( 2A - 3A \right)}

\frac{\sin 7A + \sin 3A - \sin 7A - \sin 5A}{\cos \left( - A \right) + \cos 3A - \cos 5A - \cos \left( - A \right)}

\frac{\sin 7A + \sin 3A - \sin 7A - \sin 5A}{\cos A + \cos 3A - \cos 5A - cos A}

\frac{\sin 3A - \sin 5A}{\cos 3A - \cos 5A}

\frac{2\sin \left( \frac{3A - 5A}{2} \right) \cos \left( \frac{3A + 5A}{2} \right)}{- 2\cos \left( \frac{3A + 5A}{2} \right) \cos\left( \frac{3A - 5A}{2} \right)}

\frac{\sin \left( - A \right) \cos 4A}{- \cos 4A \cos \left( - A \right)}

\frac{- \sin A \cos 4A}{- \cos 4A\cos A}

\frac{\sin A}{\cos A}

= tan A



= R.H.S.

Hence proved.

(vii)\frac{sin11AsinA+sin7Asin3A}{cos11AsinA+cos7Asin3A}=tan8A

Solution:

We have,

L.H.S. =\frac{sin11AsinA+sin7Asin3A}{cos11AsinA+cos7Asin3A}

=\frac{cos10A-cos12A+cos4A-cos10A}{sin12A-sin10A+sin10A-sin4A}

=\frac{cos4A-cos12A}{sin12A-sin4A}

=\frac{2sin\frac{16A}{2}sin\frac{8A}{2}}{2cos\frac{16A}{2}sin\frac{8A}{2}}

=\frac{2sin8Asin4A}{2cos8Asin4A}

= tan 8A

= R.H.S.

Hence proved.

(viii) \frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A \sin A + \cos 6A \cos A} = \tan 2A

Solution:

We have,

L.H.S. = \frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A sin A + \cos 6A \cos A}

On multiplying numerator and denominator by 2, we get

\frac{2\sin 3A \cos 4A - 2\sin A \cos 2A}{2\sin 4A \sin A + 2\cos 6A \cos A}

\frac{\sin \left( 3A + 4A \right) + \sin \left( 3A - 4A \right) - \sin \left( A + 2A \right) - \sin \left( A - 2A \right)}{\cos \left( 4A - A \right) - \cos \left( 4A + A \right) + \cos \left( 6A + A \right) + \cos \left( 6A - A \right)}

\frac{\sin 7A + \sin \left( - A \right) - \sin 3A - \sin \left( - A \right)}{\cos 3A - \cos 5A + \cos 7A + \cos 5A}

\frac{\sin 7A - \sin A - \sin 3A + \sin A}{\cos 3A - \cos 5A + \cos 7A + \cos 5A}



\frac{\sin 7A - \sin 3A}{\cos 3A + \cos 7A}

\frac{2\sin \left( \frac{7A - 3A}{2} \right) \cos \left( \frac{7A + 3A}{2} \right)}{2\cos \left( \frac{3A + 7A}{2} \right) \cos \left( \frac{3A - 7A}{2} \right)}

\frac{\sin 2A \cos 5A}{\cos 5A \cos \left( - 2A \right)}

\frac{\sin 2A \cos 5A}{\cos 5A \cos 2A}

= tan 2A

= R.H.S.

Hence proved.

(ix) \frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A} = \tan 5A

Solution:

We have,

L.H.S. = \frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A}

On multiplying numerator and denominator by 2, we get

\frac{2\sin A \sin 2A + 2\sin 3A \sin 6A}{2\sin A \cos 2A + 2\sin 3A \cos 6A}

\frac{\cos \left( A - 2A \right) - \cos \left( A + 2A \right) + \cos \left( 3A - 6A \right) - \cos \left( 3A + 6A \right)}{\sin \left( A + 2A \right) + \sin \left( A - 2A \right) + \sin \left( 3A + 6A \right) + \sin \left( 3A - 6A \right)}

\frac{\cos\left( - A \right) - \cos 3A + \cos \left( - 3A \right) - \cos 9A}{\sin 3A \sin\left( - A \right) + \sin 9A + \sin \left( - 3A \right)}

\frac{\cos A - \cos 3A + \cos 3A - \cos 9A}{\sin 3A - \sin A + \sin 9A - \sin 3A}

\frac{\cos A - \cos 9A}{\sin 9A - \sin A}

\frac{- 2\sin \left( \frac{A + 9A}{2} \right) \sin \left( \frac{A - 9A}{2} \right)}{2\cos \left( \frac{A + 9A}{2} \right) \sin \left( \frac{9A - A}{2} \right)}

\frac{\sin5A\cos4A}{\sin 5A \cos \left( - 4A \right)}

= tan 5A

= R.H.S.



Hence proved.

(x) \frac{\sin A + 2 \sin 3A + \sin 5A}{\sin 3A + 2 \sin 5A + \sin 7A} = \frac{\sin 3A}{\sin 5A}

Solution:

We have,

L.H.S. = \frac{\sin A + 2 \sin 3A + \sin 5A}{\sin 3A + 2 \sin 5A + \sin 7A}

\frac{\sin 5A \sin A + 2\sin 3A}{\sin 7A \sin 3A + 2\sin 5A }

\frac{2\sin \left( \frac{5A +A}{2} \right) \cos \left( \frac{5A - A}{2}\right) + 2\sin \left(3A \right)}{2\sin \left(\frac{ 7A + 3A}{2} \right) \cos \left(\frac{7A - 3A}{2} \right) + 2\sin \left( 5A \right) }

\frac{2\sin3A\cos2A+2\sin3A}{2\sin5A\cos2A + 2\sin5A}

\frac{2\sin3A(\cos2A +1)}{2\sin5A(\cos2A +1)}

= sin3A/sin5A

= R.H.S.

Hence proved.

(xi)\frac{sin(θ+Ø)-2sinθ+sin(θ-Ø)}{cos(θ+Ø)-2cosθ+cos(θ-Ø)}=tanθ

Solution:

We have,

L.H.S. =\frac{sin(θ+Ø)-2sinθ+sin(θ-Ø)}{cos(θ+Ø)-2cosθ+cos(θ-Ø)}

=\frac{sin(θ+Ø)+sin(θ-Ø)-2sinθ}{cos(θ+Ø)+cos(θ-Ø)-2cosθ}

=\frac{2sin\frac{(θ+Ø+θ-Ø)}{2}cos\frac{(θ+Ø-θ+Ø)}{2}-2sinθ}{2cos\frac{(θ+Ø+θ-Ø)}{2}cos\frac{(θ+Ø-θ+Ø)}{2}-2cosθ}

=\frac{2sin\frac{2θ}{2}cos\frac{2Ø}{2}-2sinθ}{2cos\frac{2θ}{2}cos\frac{2Ø}{2}-2cosθ}

=\frac{2sinθcosØ-2sinθ}{2cosθcosØ-2cosθ}

=\frac{2sinθ(cosØ-1)}{2cosθ(cosØ-1)}

=\frac{sinθ}{cosθ}



= tan θ

= R.H.S.

Hence proved.

Question 9. Prove that:

(i) sin α + sin β + sin γ – sin (α + β + γ) = 4 sin (α + β)/2 sin (β + γ)/2 sin (α + γ)/2

Solution:

We have,

L.H.S. = sin α + sin β + sin γ – sin (α + β + γ)

=2sin\frac{α+β}{2}cos\frac{α-β}{2}+2cos(\frac{γ+α+β+γ}{2})sin(\frac{γ-α-β-γ}{2})

=2sin\frac{α+β}{2}cos\frac{α-β}{2}+2cos(\frac{α+β+2γ}{2})sin(\frac{-(α+β)}{2})

=2sin\frac{α+β}{2}cos\frac{α-β}{2}-2cos(\frac{α+β+2γ}{2})sin(\frac{α+β}{2})

=2sin\frac{α+β}{2}(cos\frac{α-β}{2}-cos\frac{α+β+2γ}{2})

=2sin\frac{α+β}{2}(-2sin\frac{\frac{α-β+α+β+2γ}{2}}{2}sin\frac{\frac{α-β-α-β-2γ}{2}}{2})

=2sin\frac{α+β}{2}(2sin\frac{α+γ}{2}sin\frac{β+γ}{2})

= 4 sin (α + β)/2 sin (β + γ)/2 sin (α + γ)/2

= R.H.S.

Hence proved.

(ii) cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (–A + B + C) = 4 cos A cos B cos C

Solution:

We have,

L.H.S. = cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (–A + B + C)

=2cos\frac{A+B+C+A-B+C}{2}cos\frac{A+B+C-A+B-C}{2}+2cos\frac{A+B-C-A+B+C}{2}cos\frac{A+B-C+A-B-C}{2}

=2cos\frac{2A+2C}{2}cos\frac{2B}{2}+2cos\frac{2B}{2}cos\frac{2A-2C}{2}

= 2 cos (A + C) cos B + 2 cos B cos (A − C)

= 2 cos B [cos (A + C) + cos (A − C)]

= 2 cos B\left[2cos\frac{A+C+A-C}{2}cos\frac{A+C-A+C}{2}\right]

= 2 cos B [2 cos A cos C]

= 4 cos A cos B cos C

= R.H.S.

Hence proved.

Question 10. If cos A + cos B = 1/2 and sin A + sin B = 1/4, prove thattan\frac{A+B}{2}=\frac{1}{2}     .

Solution:

We have,

cos A + cos B = 1/2

sin A + sin B = 1/4

=>\frac{sinA+sinB}{cosA+cosB}=\frac{\frac{1}{4}}{\frac{1}{2}}

=>\frac{sinA+sinB}{cosA+cosB}=\frac{1}{2}

=>\frac{2sin\frac{A+B}{2}cos\frac{A-B}{2}}{2cos\frac{A+B}{2}cos\frac{A-B}{2}}=\frac{1}{2}

=>\frac{sin\frac{A+B}{2}}{cos\frac{A+B}{2}}=\frac{1}{2}

=>tan\frac{A+B}{2}=\frac{1}{2}

Hence proved.

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