Question

Prove that:$\frac{\sin (x+y)}{\sin (x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y}$

Answer 1

L.H.S$=\frac{\sin (x+y)}{\sin (x-y)}$

$\frac{\sin x\cos y+\cos x\sin y}{\sin x\cos y-\cos x\sin y}$

R.H.S$=\frac{\tan x+\tan y}{\tan x-\tan y}$

$=\frac{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}{\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}}$

$=\frac{\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}}{\frac{\sin x\cos y-\cos x\sin y}{\cos x\cos y}}$

$=\frac{\sin x\cos y+\cos x\sin y}{\sin x\cos y-\cos x\sin y}$

$=$L.H.S

Hence L.H.S$=$R.H.S

Hence proved.