Given: sin α + sin β = a & cos α + cos β = b

Proof:

sin α + sin β = a

Squaring both sides, we get

(sin α + sin β)² = a²

⇒ sin² α + sin² β + 2 sin α sin β = a² ……(1)

cos α + cos β = b

Squaring both sides, we get

(cos α + cos β)² = a²

⇒ cos² α + cos² β + 2 cos α cos β = b² ………(2)

Adding equation 1 and 2, we get

sin² α + sin² β + 2 sin α sin β + cos² α + cos² β + 2 cos α cos β = a² + b²

⇒ sin² α + cos² α + sin² β + cos² β + 2 sin α sin β + 2 cos α cos β = a² + b²

⇒ 1 + 1 + 2 sin α sin β + 2 cos α cos β = a² + b²

{∵ sin² x + cos² x = 1}

⇒ 2 + 2 sin α sin β + 2 cos α cos β = a² + b²

⇒ 2(sin α sin β + cos α cos β) = a² + b² – 2

We know,

sin A sin B + cos A cos B = cos (A – B)

Therefore,

Hence Proved