Question

Prove that $\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$

Answer 1

To Prove:

$\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$

We know that: $tanA=\frac{\sin A}{\cos A}$

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

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

Dividing Numeator and denominator by $\left(\cos x\cos y\right)$:

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

$\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$

Hence Proved.