Question

Prove the following:
$\frac{\text{sin x - sin y}}{\text{cos x + cos y}}=\text{tan}\left(\frac{x-y}{2}\right)$

Answer 1

We have

L.H.S. = $\frac{\text{sin x - sin y}}{\text{cos x + cos y}}$

=$\frac{\text{2 cos}\left(\frac{x+y}{2}\right)\text{sin}\left(\frac{x-y}{2}\right)}{\text{2 cos}\left(\frac{x+y}{2}\right)\text{cos}\left(\frac{x-y}{2}\right)}$

$\left[\begin{array}{c}\because \text{sin C - sin D}=\text{2 cos}\left(\frac{C+D}{2}\right)\text{sin}\left(\frac{C-D}{2}\right)\\ \text{cos C + cos D}=\text{2 cos}\left(\frac{C+D}{2}\right)\text{cos}\left(\frac{C-D}{2}\right)\end{array}\right]$

$=\frac{\text{sin}\frac{x-y}{2}}{\text{cos}\frac{x-y}{2}}=\text{tan}\frac{x-y}{2}$

= R.H.S