Question

How do you solve $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\times \tan y}$?

Answer 1

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

$\sin (x+y)=\sin x\times \cos y+\cos x\times \sin y$

$\cos (x+y)=\cos x\times \cos y–\sin x\times s\mathrm{in}y$

Taking the L.H.S

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

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

Divide all the terms by $\cos x\times \cos y$

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

$\Rightarrow$ $\tan \left(x+y\right)=\frac{(\tan x+\tan y)}{(1–\tan x\times \tan y)}$

$\Rightarrow$ $\tan \left(x+y\right)=$R.H.S

$\Rightarrow$ L.H.S$=$R.H.S

Hence, it is proven that $\tan \left(x+y\right)=\frac{(\tan x+\tan y)}{(1–\tan x\times \tan y)}$