Question

sin x + sin y / sin x - sin y = tan x + y / 2 cot x - y / 2​

Answer 1

Complete Question :

Prove that : ( sin x + sin y ) / ( sin x - sin y ) = tan [ ( x + y )/ 2 ] .  cot[ ( x - y ) / 2​ ]

Answer :

( sin x + sin y ) / ( sin x - sin y ) = tan [ ( x + y )/ 2 ] .  cot[ ( x - y ) / 2​ ]

Consider LHS

=  ( sin x + sin y ) / ( sin x - sin y )

Using Sum - to - Product identities

• $\rm sin \ x+sin \ y =2sin\Bigg(\dfrac{x+y}{2}\Bigg).cos\Bigg(\dfrac{x-y}{2}\Bigg)$

• $\rm sin \ x-sin \ y =2cos\Bigg(\dfrac{x+y}{2}\Bigg).sin\Bigg(\dfrac{x-y}{2}\Bigg)$

$\sf =\dfrac{2sin\Bigg(\dfrac{x+y}{2}\Bigg).cos\Bigg(\dfrac{x-y}{2}\Bigg)}{2cos\Bigg(\dfrac{x+y}{2}\Bigg).sin\Bigg(\dfrac{x-y}{2}\Bigg)}$

$\sf =\dfrac{sin\Bigg(\dfrac{x+y}{2}\Bigg).cos\Bigg(\dfrac{x-y}{2}\Bigg)}{cos\Bigg(\dfrac{x+y}{2}\Bigg).sin\Bigg(\dfrac{x-y}{2}\Bigg)}$

It can be written as

$\sf =\dfrac{sin\Bigg(\dfrac{x+y}{2}\Bigg)}{cos\Bigg(\dfrac{x+y}{2}\Bigg)} \times \dfrac{cos\Bigg(\dfrac{x-y}{2}\Bigg)}{sin\Bigg(\dfrac{x-y}{2}\Bigg)}$

Using sin ∅ / cos ∅ = tan ∅ and cos ∅ / sin ∅ = cot ∅ we get ,

$\sf =tan \Bigg(\dfrac{x+y}{2}\Bigg) . cot \Bigg(\dfrac{x-y}{2}\Bigg) \\\\\\ \sf = RHS$

Hence Proved.