Given:
[tex]\frac{2\tan x}{1+\tan^2x}=\sin 2x[/tex]
Take the left-hand side of the equation,
[tex]\begin{gathered} \text{LHS}=\frac{2\tan x}{1+\tan^2x} \\ \text{Use the identity: 1+tan}^2x=sex^2x \\ =\frac{2\tan x}{\sec^2x} \\ =\frac{2\frac{\sin x}{\cos x}}{\frac{1}{\cos^2x}}\ldots.\ldots\text{.. Since tanx=}\frac{sinx}{\cos x},\text{secx}=\frac{1}{\cos x} \\ =2\frac{\sin x}{\cos x}\times\cos ^2x \\ =2\sin x\times\cos x \\ =\sin 2x\ldots\ldots...\ldots\text{ Since }\sin 2\text{x=2}\sin x\cdot\cos x \\ =\text{ Left hand side} \end{gathered}[/tex]
Hence, it is proved that,
[tex]\frac{2\tan x}{1+\tan^2x}=\sin 2x[/tex]
