sin C + sin D = 2 sin [(C + D)/2] cos [(C - D)/2]

sin C - sin D = 2 cos [(C + D)/2] sin [(C - D)/2]

cos C + cos D = 2 cos [(C + D)/2] cos [(C - D)/2]

cos C - cos D = -2 sin [(C + D)/2] sin [(C - D)/2]

**Problem 1 :**

Prove that

sin(4A-2B) + sin(4B-2A) /cos(4A-2B) + cos(4B-2A) = tan(A+B)

**Solution :**

sin(4A-2B) + sin(4B-2A)

= 2sin(A + B)cos(3A+3B)

cos(4A-2B) + cos(4B-2A)

= 2cos(A+B)cos(3A+3B)

Then,

sin(4A-2B) + sin(4B-2A) /cos(4A-2B) + cos(4B-2A) :

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

= tan(A+B)

Hence proved.

**Problem 2 :**

Prove that :

cot(A+15) - tan(A-15) = 4cos2A/(1 + 2sin2A)

**Solution :**

cot(A+15) = cos(A+15) / sin(A+15)

tan(A+15) = sin(A-15) / cos(A-15)

Then,

cot(A+15) - tan(A-15) :

= [cos(A+15)/sin(A-15)] - [sin(A-15)/cos(A-15)]

= [cos (A+15)cos(A-15) - sin(A-15)sin(A+15)] / sin(A+15)cos (A-15) -------(1)

cos(A+15)cos(A-15) = (1/2)[2cos(A+15) cos (A-15)]

= (1/2)[cos2A + cos30]

sin(A-15)sin(A+15) = (-1/2)[-2 sin (A-15) sin (A+15)]

= (-1/2)[cos2A - cos30]

**Numerator of (1) :**

cos(A+15)cos(A-15) - sin(A-15)sin(A+15) :

= (1/2)[cos2A + cos30] - (-1/2)[cos2A - cos30]

= (1/2)[2cos2A]

= cos2A

**Denominator of (1) :**

sin (A+15)cos(A-15) :

= (1/2)[2sin(A+15)cos(A-15)]

= (1/2)[sin2A + sin30]

= (1/2)[sin2A + (1/2)]

= (1/4)[2sin2A + 1]

Then,

(1)-----> = cos 2A/[(1/4)(2sin2A + 1)]

= 4cos2A / (2sin2A + 1)

Hence it is proved.

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