#### Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 31

$\left [ -\pi,\pi \right ]$

Hint:

You must know the rules of inverse trigonometric function.

Given:

$y=2\tan^{-1}x+\sin^{-1}\left ( \frac{2x}{1+x^{2}} \right )$

Solution:

$y= 2\tan^{-1}x+\sin^{-1}\left ( \frac{2x}{1+x^{2}} \right )$

Let, $x=\tan \tan\theta$

$\therefore y=2\tan\theta+\left ( \frac{2\tan\tan\theta}{1+\theta} \right )$

$Y=2\theta+\sin^{-1}\left ( \sin2\theta \right )$                                                                                                $\left [ \therefore \sin2\theta=\left ( \frac{2\tan\theta}{1+\tan^{2}\theta} \right ) \right ]$

$\Rightarrow y=2\theta+2\theta$

$\Rightarrow y=4\theta$                                                                                 $\left [ \therefore \theta=\tan^{-1} \left ( x \right )\right ]$

$\Rightarrow y=4\tan^{-1}x$

$\because -\frac{\pi}{2}< \tan^{-1}x< \frac{\pi}{2}$

$\because -\frac{4\pi}{2}<4 \tan^{-1}x< \frac{4\pi}{2}$

$\Rightarrow -2\pi< 4\tan^{-1}x< 2\pi$

$\Rightarrow -2\pi< y< 2\pi$                                                                                     $\left [ y=4\tan^{-1}x \right ]$

Range of $y$ is $\left [ -2\pi,2\pi \right ]$