Question

How do you prove $\frac{2\tan x}{1+{\tan }^{2}x}=\sin 2x$?

Answer 1

Proof of given relation:

$\frac{2\tan x}{1+{\tan }^{2}x}=\sin 2x$

Here we have $LHS=\frac{2\tan x}{1+{\tan }^{2}x}$

Therefore,

$\frac{2\tan x}{1+{\tan }^{2}x}=\frac{2{\displaystyle \frac{\sin x}{\cos x}}}{se{c}^{2}x}$ $[\because tanx=\frac{sinx}{cosx},1+{\tan }^{2}x={\sec }^{2}x]$

$=2\frac{\sin x}{\cos x}\times {\cos }^{2}x$ $\left[\because secx=\frac{1}{cosx}\right]$

$=2\sin x\cos x$

$=\sin 2x$ $\left[\because sin2x=2sinxcosx\right]$

$=RHS$

Hence, $\frac{\mathbf{2}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{x}}{\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{x}}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{x}$ is proved.