Question

sin (x+y)/sin (x-y)=tanx+tany/tanx-tany proof that

Answer 1

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Answer 2

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

Step-by-step explanation:

To prove, $\dfrac{\sin (x+y}{\sin (x-y)}=\dfrac{\tan x+\tan y}{\tan x-\tan y}$.

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

Using trigonometric identity,

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

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

$=\dfrac{\sin x.\cos y+\cos x.\sin y}{\sin x.\cos y-\cos x.\sin y}$

Dividing nemerator and denominator by $\cos x.\cos y$, we get

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

$=\dfrac{\tan x+\tan y}{\tan x-\tan y}$

= R.H.S, proved.

Hence, $\dfrac{\sin (x+y}{\sin (x-y)}=\dfrac{\tan x+\tan y}{\tan x-\tan y}$