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

$\left [ \frac{\pi}{4} ,\frac{3\pi}{4}\right ]$

Hint:

You must know the value of trigonometric and inverse trigonometric function.

Given:

Range of $\sin^{-1}x+\cos^{-1}x+\tan^{-1}x$

Solution:

$\sin^{-1}x+\cos^{-1}x+\tan^{-1}x=f\left ( x \right )$

$\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$     [Inverse trigonometric function]

So, we get

$\Rightarrow$         $\frac{\pi}{2}+\tan^{-1}x=f\left ( x \right )$

Now, most of us apply the range of $\tan^{-1}x$ as  $\left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$

So, range of  $f\left ( x \right )=\left [ 0,\pi \right ]$

But, we know that domain of $f\left ( x \right )$ is $\left [ -1,1 \right ]$

Hence, term $\tan^{-1}x$ does not hold this value$\frac{-\pi}{2}$  and $\frac{\pi}{2}$

So new range becomes,

$\Rightarrow f\left ( -1 \right )=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$

$\Rightarrow f\left ( 1 \right )=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}$

So, we get  $f\left ( x \right )\epsilon \left [ \frac{\pi}{4},\frac{3\pi}{4} \right ]$