Question

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

Answer 1

RHS $\frac{\tan x+\tan y}{\tan x-\tan y}$

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

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

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

$=\frac{\sin (x+y)}{\sin (x-y)}$

Hence proved.