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

### (i) sin 12Î¸ + sin 4Î¸

Solution:

We know, sin A + sin B = 2 sin (A+B)/2 cos (Aâ€“B)/2

sin 12Î¸ + sin 4Î¸ = 2 sin (12Î¸ + 4Î¸)/2 cos (12Î¸ â€“ 4Î¸)/2

= 2 sin 16Î¸/2 cos 8Î¸/2

= 2 sin 8Î¸ cos 4Î¸

### (ii) sin 5Î¸ â€“ sin Î¸

Solution:

We know, sin A â€“ sin B = 2 cos (A+B)/2 sin (Aâ€“B)/2

sin 5Î¸ â€“ sin Î¸ = 2 cos (5Î¸ + Î¸)/2 sin (5Î¸ â€“ Î¸)/2

= 2 cos 6Î¸/2 sin 4Î¸/2

= 2 cos 3Î¸ sin 2Î¸

### (iii) cos 12Î¸ + cos 8Î¸

Solution:

We know, cos A + cos B = 2 cos (A+B)/2 cos (Aâ€“B)/2

cos 12Î¸ + cos 8Î¸ = 2 cos (12Î¸ + 8Î¸)/2 cos (12Î¸ â€“ 8Î¸)/2

= 2 cos 20Î¸/2 cos 4Î¸/2

= 2 cos 10Î¸ cos 2Î¸

### (iv) cos 12Î¸ â€“ cos 4Î¸

Solution:

We know, cos A â€“ cos B = â€“2 sin (A+B)/2 sin (Aâ€“B)/2

cos 12Î¸ â€“ cos 4Î¸ = â€“2 sin (12Î¸ + 4Î¸)/2 sin (12Î¸ â€“ 4Î¸)/2

= â€“2 sin 16Î¸/2 sin 8Î¸/2

= â€“2 sin 8Î¸ sin 4Î¸

### (v) sin 2Î¸ + cos 4Î¸

Solution:

sin 2Î¸ + cos 4Î¸ = sin 2Î¸ + sin (90o â€“ 4Î¸)

We know, sin A + sin B = 2 sin (A+B)/2 cos (Aâ€“B)/2

sin 2Î¸ + sin (90o â€“ 4Î¸) = 2 sin (2x + 90o â€“ 4Î¸)/2 cos (2Î¸ â€“ 90o + 4Î¸)/2

= 2 sin (90o â€“ 2Î¸)/2 cos (6Î¸ â€“ 90o)/2

= 2 sin (45o â€“ Î¸) cos (3Î¸ â€“ 45o)

= 2 sin (45o â€“ Î¸) cos [â€“(45o â€“ 3Î¸)]

= 2 sin (45o â€“ Î¸) cos (45o â€“ 3Î¸)

= 2 sin (Ï€/4 â€“ Î¸) cos (Ï€/4 â€“ 3Î¸)

### (i) sin 38Â° + sin 22Â° = sin 82Â°

Solution:

Given, L.H.S. = sin 38Â° + sin 22Â°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (Aâ€“B)/2

sin 38Â° + sin 22Â° = 2 sin (38o + 22o)/2 cos (38o â€“ 22o)/2

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

= 2 sin 30o cos 8o

= 2 Ã— (1/2) Ã— cos 8o

= cos 8o

= cos (90Â° â€“ 82Â°)

= sin 82Â°

= R.H.S.

Hence proved.

### (ii) cos 100Â° + cos 20Â° = cos 40Â°

Solution:

Given, L.H.S. = cos 100Â° + cos 20Â°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (Aâ€“B)/2

cos 100Â° + cos 20Â° = 2 cos (100o + 20o)/2 cos (100o â€“ 20o)/2

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

= 2 cos 60o cos 40o

= 2 Ã— (1/2) Ã— cos 40o

= cos 40o

= R.H.S.

Hence Proved.

### (iii) sin 50Â° + sin 10Â° = cos 20Â°

Solution:

Given, L.H.S. = sin 50Â° + sin 10Â°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (Aâ€“B)/2.

sin 50Â° + sin 10Â° = 2 sin (50o + 10o)/2 cos (50o â€“ 10o)/2

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

= 2 sin 30o cos 20o

= 2 Ã— (1/2) Ã— cos 20o

= cos 20o

= R.H.S

Hence Proved.

### (iv) sin 23Â° + sin 37Â° = cos 7Â°

Solution:

Given L.H.S. = sin 23Â° + sin 37Â°.

We know sin A + sin B = 2 sin (A+B)/2 cos (Aâ€“B)/2

sin 23Â° + sin 37Â° = 2 sin (23o + 37o)/2 cos (23o â€“ 37o)/2

= 2 sin 60o/2 cos (â€“14o/2)

= 2 sin 30o cos (â€“7o)

= 2 Ã— (1/2) Ã— cos 7o

= cos 7o

= R.H.S.

Hence Proved.

### (v) sin 105Â° + cos 105Â° = cos 45Â°

Solution:

Given, L.H.S. = sin 105Â° + cos 105Â°

sin 105Â° + cos 105Â° = sin 105o + sin (90o â€“ 105o)

= sin 105o + sin (â€“15o)

= sin 105o â€“ sin 15o

We know, sin A â€“ sin B = 2 cos (A+B)/2 sin (Aâ€“B)/2

sin 105o â€“ sin 15o = 2 cos (105o + 15o)/2 sin (105o â€“ 15o)/2

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

= 2 cos 60o sin 45o

= 2 Ã— (1/2) Ã— (1/âˆš2)

= 1/âˆš2

= cos 45o

= R.H.S.

Hence Proved.

### (vi) sin 40Â° + sin 20Â° = cos 10Â°

Solution:

Given, L.H.S. = sin 40Â° + sin 20Â°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (Aâ€“B)/2

sin 40Â° + sin 20Â° = 2 sin (40o + 20o)/2 cos (40o â€“ 20o)/2

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

= 2 sin 30o cos 10o

= 2 Ã— (1/2) Ã— cos 10o

= cos 10o

= R.H.S.

Hence Proved.

### (i) cos 55Â° + cos 65Â° + cos 175Â° = 0

Solution:

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

We know, cos A + cos B = 2 cos (A+B)/2 cos (Aâ€“B)/2

cos 55Â° + cos 65Â° + cos 175Â° = 2 cos (55o + 65o)/2 cos (55o â€“ 65o) + cos (180o â€“ 5o)

= 2 cos 120o/2 cos (â€“10o)/2 â€“ cos 5o

= 2 cos 60Â° cos (â€“5Â°) â€“ cos 5Â°

= 2 Ã— (1/2) Ã— cos 5o â€“ cos 5o

= cos 5o â€“ cos 5o

= 0

= R.H.S.

Hence Proved.

### (ii) sin 50Â° â€“ sin 70Â° + sin 10Â° = 0

Solution:

Given, L.H.S. = sin 50Â° â€“ sin 70Â° + sin 10Â°.

We know, sin A â€“ sin B = 2 cos (A+B)/2 sin (Aâ€“B)/2

sin 50Â° â€“ sin 70Â° + sin 10Â° = 2 cos (50o + 70o)/2 sin (50o â€“ 70o) + sin 10o

= 2 cos 120o/2 sin (â€“20o)/2 + sin 10o

= 2 cos 60o (â€“sin 10o) + sin 10o

= 2 Ã— (1/2) Ã— (â€“sin 10o) + sin 10o

= 0

= R.H.S.

Hence Proved.

### (iii) cos 80Â° + cos 40Â° â€“ cos 20Â° = 0

Solution:

Given L.H.S. = cos 80Â° + cos 40Â° â€“ cos 20Â°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (Aâ€“B)/2

cos 80Â° + cos 40Â° â€“ cos 20Â° = 2 cos (80o + 40o)/2 cos (80o â€“ 40o) â€“ cos 20o

= 2 cos 120o/2 cos 40o/2 â€“ cos 20o

= 2 cos 60Â° cos 20o â€“ cos 20Â°

= 2 Ã— (1/2) Ã— cos 20o â€“ cos 20o

= 0

= R.H.S.

Hence Proved.

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

Solution:

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

We know, cos A + cos B = 2 cos (A+B)/2 cos (Aâ€“B)/2.

cos 20Â° + cos 100Â° + cos 140Â° = 2 cos (20o + 100o)/2 cos (20o â€“ 100o) + cos (80o â€“ 40o)

= 2 cos 120o/2 cos (â€“80o)/2 â€“ cos 40o

= 2 cos 60Â° cos (â€“40Â°) â€“ cos 40Â°

= 2 Ã— (1/2) Ã— cos 40o â€“ cos 40o

= 0

= R.H.S.

Hence Proved.

### (v) sin 5Ï€/18 â€“ cos 4Ï€/9 = âˆš3 sin Ï€/9

Solution:

Given, L.H.S. = sin 5Ï€/18 â€“ cos 4Ï€/9

= sin 5Ï€/18 â€“ sin (Ï€/2 â€“ 4Ï€/9)

= sin 5Ï€/18 â€“ sin (9Ï€ â€“ 8Ï€)/18

= sin 5Ï€/18 â€“ sin Ï€/18

We know, sin A â€“ sin B = 2 cos (A+B)/2 sin (Aâ€“ B)/2

= 2 cos (6Ï€/36) sin (4Ï€/36)

= 2 cos Ï€/6 sin Ï€/9

= 2 cos 30o sin Ï€/9

= 2 Ã— (âˆš3/2) Ã— sin Ï€/9

= âˆš3 sin Ï€/9

= R.H.S.

Hence Proved.

### (vi) cos Ï€/12 â€“ sin Ï€/12 = 1/âˆš2

Solution:

Given, cos Ï€/12 â€“ sin Ï€/12 = sin (Ï€/2 â€“ Ï€/12) â€“ sin Ï€/12

= sin (6Ï€ â€“ 5Ï€)/12 â€“ sin Ï€/12

= sin 5Ï€/12 â€“ sin Ï€/12

We know, sin A â€“ sin B = 2 cos (A+B)/2 sin (Aâ€“B)/2

= 2 cos (6Ï€/24) sin (4Ï€/24)

= 2 cos Ï€/4 sin Ï€/6

= 2 cos 45o sin 30o

= 2 Ã— (1/âˆš2) Ã— (1/2)

= 1/âˆš2

= R.H.S.

Hence Proved.

### (vii) sin 80Â° â€“ cos 70Â° = cos 50Â°

Solution:

We have, sin 80Â° = cos 50Â° + cos 70o

Here, R.H.S. = cos 50Â° + cos 70o

We know,

cos A + cos B = 2 cos (A+B)/2 cos (Aâ€“B)/2

cos 50Â° + cos 70o = 2 cos (50o + 70o)/2 cos (50o â€“ 70o)/2

= 2 cos 120o/2 cos (â€“20o)/2

= 2 cos 60o cos (â€“10o)

= 2 Ã— (1/2) Ã— cos 10o

= cos 10o

= cos (90Â° â€“ 80Â°)

= sin 80Â°

= L.H.S.

Hence Proved.

### (viii) sin 51Â° + cos 81Â° = cos 21Â°

Solution:

Given, L.H.S. = sin 51Â° + cos 81Â°

= sin 51o + sin (90o â€“ 81o)

= sin 51o + sin 9o

We know, sin A + sin B = 2 sin (A+B)/2 cos (Aâ€“B)/2

sin 51o + sin 9o = 2 sin (51o + 9o)/2 cos (51o â€“ 9o)/2

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

= 2 sin 30o cos 21o

= 2 Ã— (1/2) Ã— cos 21o

= cos 21o

= R.H.S.

Hence Proved.

### (i) cos (3Ï€/4 + x) â€“ cos (3Ï€/4 â€“ x) = â€“âˆš2 sin x

Solution:

Given, L.H.S. = cos (3Ï€/4 + x) â€“ cos (3Ï€/4 â€“ x)

We know, cos A â€“ cos B = â€“2 sin (A+B)/2 sin (Aâ€“B)/2

cos (3Ï€/4 + x) â€“ cos (3Ï€/4 â€“ x) = â€“2 sin (3Ï€/4 + x + 3Ï€/4 â€“ x)/2 sin (3Ï€/4 + x â€“ 3Ï€/4 + x)/2

= â€“2 sin (6Ï€/4)/2 sin 2x/2

= â€“2 sin 6Ï€/8 sin x

= â€“2 sin 3Ï€/4 sin x

= â€“2 sin (Ï€ â€“ Ï€/4) sin x

= â€“2 sin Ï€/4 sin x

= â€“2 Ã— (1/âˆš2) Ã— sin x

= â€“âˆš2 sin x

= R.H.S.

Hence proved.

### (ii) cos (Ï€/4 + x) + cos (Ï€/4 â€“ x) = âˆš2 cos x

Solution:

Given, L.H.S. = cos (Ï€/4 + x) + cos (Ï€/4 â€“ x)

We know, cos A + cos B = 2 cos (A+B)/2 cos (Aâ€“B)/2

cos (Ï€/4 + x) + cos (Ï€/4 â€“ x) = 2 cos (Ï€/4 + x + Ï€/4 â€“ x)/2 cos (Ï€/4 + x â€“ Ï€/4 + x)/2

= 2 cos (2Ï€/4)/2 cos 2x/2

= 2 cos 2Ï€/8 cos x

= 2 sin Ï€/4 cos x

= 2 Ã— (1/âˆš2) Ã— cos x

= âˆš2 cos x

= R.H.S.

Hence proved.

### (i) sin 65o + cos 65o = âˆš2 cos 20o

Solution:

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

= sin 65o + sin (90o â€“ 65o)

= sin 65o + sin 25o

We know, sin A + sin B = 2 sin (A+B)/2 cos (Aâ€“B)/2

sin 65o + sin 25o = 2 sin (65o + 25o)/2 cos (65o â€“ 25o)/2

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

= 2 sin 45o cos 20o

= 2 Ã— (1/âˆš2) Ã— cos 20o

= âˆš2 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 â€“ 77o)

= sin 47o + sin 13o

We know, sin A + sin B = 2 sin (A+B)/2 cos (Aâ€“B)/2

sin 47o + sin 13o = 2 sin (47o + 13o)/2 cos (47o â€“ 13o)/2

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

= 2 sin 30o cos 17o

= 2 Ã— (1/2) Ã— cos 17o

= cos 17o

= R.H.S.

Hence proved.

### (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 (Aâ€“B)/2

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

= [2 cos (5A+3A)/2 cos (5Aâ€“3A)/2] + [2 cos (15A+7A)/2 cos (15Aâ€“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 (Aâ€“B)/2

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

= [2 cos (3A+A)/2 cos (3Aâ€“A)/2] + [2 cos (7A+5A)/2 cos (7Aâ€“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 (6Aâ€“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 (Aâ€“B)/2

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

= [2 sin (2A+A)/2 cos (2Aâ€“A)/2] + [2 sin (5A+4A)/2 cos (5Aâ€“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 â€“ 3A/2)/2]

= 2 cos A/2 [2 sin ((9A+3A)/2)/2 cos ((9Aâ€“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 â€“ sin A = 4 sin A cos A/2 cos 3A/2

Solution:

Given, L.H.S. = sin 3A + sin 2A â€“ sin A

= (sin 3A â€“ sin A) + sin 2A

We know, sin A â€“ sin B = 2 cos (A+B)/2 sin (Aâ€“B)/2

= (sin 3A â€“ sin A) + sin 2A

= 2 cos (3A + A)/2 sin (3A â€“ 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 â€“ cos 140o cos 200o = â€“ 3/4

Solution:

Given L.H.S. = cos 20o cos 100o + cos 100o cos 140o â€“ cos 140o cos 200o

= 1/2 [2 cos 100o cos 20o + 2 cos 140o cos 100o â€“ 2 cos 200o cos 140o]

We know that, 2 cos A cos B = cos (A+B) + cos (Aâ€“B)

= 1/2 [cos (100o + 20o) + cos (100o â€“ 20o) + cos (140o + 100o) + cos (140o â€“ 100o) â€“ cos (200o + 140o) â€“ cos (200o â€“ 140o)]]

= 1/2 [cos 120o + cos 80o + cos 240o + cos 40o â€“ cos 340o â€“ cos 60o]

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

= 1/2 [â€“sin 30o + cos 80o â€“ cos 60o + cos 40o â€“ cos 20o â€“ cos 60o]

= 1/2 [â€“sin 30o + cos 80o + cos 40o â€“ cos 20o â€“ 2 cos 60o]

= 1/2 [â€“sin 30o + 2 cos (80o+40o)/2 cos (80oâ€“40o)/2 â€“ cos 20o â€“ 2 Ã— 1/2]

= 1/2 [â€“sin 30o + 2 cos 120o/2 cos 40o/2 â€“ cos 20o â€“ 1]

= 1/2 [â€“sin 30o + 2 cos 60o cos 20o â€“ cos 20o â€“ 1]

= 1/2 [â€“1/2 + 2Ã—(1/2)Ã—cos 20o â€“ cos 20o â€“ 1]

= 1/2 [â€“1/2 + cos 20o â€“ cos 20o â€“ 1]

= 1/2 [â€“1/2 â€“1]

= 1/2 [â€“3/2]

= â€“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 (Aâ€“B) â€“ cos (A+B)

= 1/2 [cos (7x/2 â€“ x/2) â€“ cos (7x/2 + x/2) + cos (11x/2 â€“ 3x/2) â€“ cos (11x/2 + 3x/2)]

= 1/2 [cos (7xâ€“x)/2 â€“ cos (7x+x)/2 + cos (11xâ€“3x)/2 â€“ cos (11x+3x)/2]

= 1/2 [cos 6x/2 â€“ cos 8x/2 + cos 8x/2 â€“ cos 14x/2]

= 1/2 [cos 3x â€“ cos 7x]

= â€“1/2 [cos 7x â€“ cos 3x]

= â€“1/2 [â€“2 sin (7x+3x)/2 sin (7xâ€“3x)/2]

= â€“1/2 [â€“2 sin 10x/2 sin 4x/2]

= â€“1/2 [â€“2 sin 5x sin 2x]

= â€“2/â€“2 sin 5x sin 2x

= sin 2x sin 5x

= R.H.S.

Hence proved.

### (vii) cos x cos x/2 â€“ cos 3x cos 9x/2 = sin 4x sin 7x/2

Solution:

Given L.H.S. = cos x cos x/2 â€“ cos 3x cos 9x/2

= 1/2 [2 cos x cos x/2 â€“ 2 cos 9x/2 cos 3x]

We know, 2 cos A cos B = cos (A+B) + cos (Aâ€“B)

= 1/2 [cos (x + x/2) + cos (x â€“ x/2) â€“ cos (9x/2 + 3x) â€“ cos (9x/2 â€“ 3x)]

= 1/2 [cos (2x+x)/2 + cos (2xâ€“x)/2 â€“ cos (9x+6x)/2 â€“ cos (9xâ€“6x)/2]

= 1/2 [cos 3x/2 + cos x/2 â€“ cos 15x/2 â€“ cos 3x/2]

= 1/2 [cos x/2 â€“ cos 15x/2]

= â€“ 1/2 [cos 15x/2 â€“ cos x/2]

= â€“ 1/2 [â€“2 sin (15x/2 + x/2)/2 sin (15x/2 â€“ x/2)/2]

= -1/2 [â€“2 sin (16x/2)/2 sin (14x/2)/2]

= -1/2 [â€“2 sin 16x/4 sin 7x/2]

= â€“ 1/2 [â€“2 sin 4x sin 7x/2]

= â€“2/â€“2 [sin 4x sin 7x/2]

= sin 4x sin 7x/2

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= cot A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= cot 8A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

=

=

=

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

=

=

=

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

=

=

=

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= tan 3A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

=

=

=

=

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= cot 3A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= tan 6A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= cot 6A

= R.H.S.

Hence proved.

### (vi)

Solution:

We have,

L.H.S. =

Multiplying numerator and denominator by 2, we get

= tan A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

=

=

=

=

= tan 8A

= R.H.S.

Hence proved.

### (viii)

Solution:

We have,

L.H.S. =

On multiplying numerator and denominator by 2, we get

= tan 2A

= R.H.S.

Hence proved.

### (ix)

Solution:

We have,

L.H.S. =

On multiplying numerator and denominator by 2, we get

= tan 5A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= sin3A/sin5A

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

=

=

=

=

=

=

= tan Î¸

= R.H.S.

Hence proved.

### (i) sin Î± + sin Î² + sin Î³ â€“ sin (Î± + Î² + Î³) = 4 sin (Î± + Î²)/2 sin (Î² + Î³)/2 sin (Î± + Î³)/2

Solution:

We have,

L.H.S. = sin Î± + sin Î² + sin Î³ â€“ sin (Î± + Î² + Î³)

=

=

=

=

=

=

= 4 sin (Î± + Î²)/2 sin (Î² + Î³)/2 sin (Î± + Î³)/2

= R.H.S.

Hence proved.

### (ii) cos (A + B + C) + cos (A â€“ B + C) + cos (A + B â€“ C) + cos (â€“A + B + C) = 4 cos A cos B cos C

Solution:

We have,

L.H.S. = cos (A + B + C) + cos (A â€“ B + C) + cos (A + B â€“ C) + cos (â€“A + B + C)

=

=

= 2 cos (A + C) cos B + 2 cos B cos (A âˆ’ C)

= 2 cos B [cos (A + C) + cos (A âˆ’ C)]

= 2 cos B

= 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 that.

Solution:

We have,

cos A + cos B = 1/2

sin A + sin B = 1/4

=>

=>

=>

=>

=>

Hence proved.

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