If f(x)=

Given:

$f(x)=\frac{2 x}{1+x^{2}}$

Thus,

$f(\tan \theta)=\frac{2(\tan \theta)}{1+\tan ^{2} \theta}$

$=\frac{2 \times \frac{\sin \theta}{\cos \theta}}{1+\left(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}\right)}$

$=\frac{2 \sin \theta}{\cos \theta} \times \frac{\cos ^{2} \theta}{\cos ^{2} \theta+\sin ^{2} \theta}$

$=\frac{2 \sin \theta \cos \theta}{1} \quad\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]$

$=\sin 2 \theta \quad[\because 2 \sin \theta \cos \theta=\sin 2 \theta]$

Hence, $f(\tan \theta)=\sin 2 \theta$.