# If $|\;x\;|\leq 1,then\;2\tan^{-1}x+\sin^{-1}\frac{2x}{1+x^2}$ is equal to:

$(A)\quad4\tan^{-1}x\quad(B)\quad 0\quad(C)\quad\frac{1}{2}\quad(D)\quad \pi$

Toolbox:
• $$\large\frac{2tan\theta}{1+tan^2\theta}=sin2\theta$$
Ans - (A) $$4tan^{-1}x$$

put $$x=tan\theta \Rightarrow \theta=tan^{-1}x$$
$$\large\frac{2x}{1+x^2}=\large\frac{2tan\theta}{1+tan^2\theta}=sin2\theta$$
$$\Rightarrow\:sin^{-1}\large\frac{2x}{1+x^2}=sin^{-1}sin2\theta=2\theta$$

Substituting the value of x, the given expression becomes
$$2tan^{-1}tan\theta+sin^{-1}\large\frac{2tan\theta}{1+tan^\theta}$$

$$\Rightarrow 2\theta+2\theta=4\theta$$
$$=4tan^{-1}x$$

edited Mar 16, 2013