Question

If $\frac{sin(x+y)}{sin(x-y)}=\frac{a+b}{a-b},$, then show that $\frac{tanx}{tany}=\frac{a}{b},$.

Answer 1

Given:- $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$

To Prove:- $\frac{\tan x}{\tan y}=\frac{a}{b}$

Proof:-

Using trigonomatric identity:

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

$\sin (A-B)=\sin A\cos B-\cos A\sin B$

Therefore,

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

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

Comparing L.H.S. and R.H.S., we have

$a=\sin x\cos y$

$b=\cos x\sin y$

$\therefore \frac{a}{b}=\frac{\sin x\cos y}{\cos x\sin y}=\tan x\cot y=\frac{\tan x}{\tan y}$

Hence proved.