Show that \(f(\tan\theta)=\sin2\theta\)

Solution

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

Put \(x=\tan\theta\):

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

Using $$ 1+\tan^2\theta=\sec^2\theta $$

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

$$ = 2\tan\theta\cos^2\theta $$

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

$$ = 2\sin\theta\cos\theta $$

Using $$ \sin2\theta=2\sin\theta\cos\theta $$

Therefore, $$ f(\tan\theta)=\sin2\theta $$

Hence, $$ \boxed{f(\tan\theta)=\sin2\theta} $$