#### Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 6

$\frac{3\pi}{4}$

Hint:

You must know the value of inverse trigonometric function.

Given:

$\tan^{-1}2+\tan^{-1}3$

Solution:

$\tan^{-1}2+\tan^{-1}3$

$\Rightarrow$         $=\left [ \frac{\pi}{2}-\cot^{-1} 2\right ]+\left [ \frac{\pi}{2}-\cot^{-1} 3\right ]$

$\Rightarrow$         $=\frac{\pi}{2}+\frac{\pi}{2}-\left [ \cot^{-1}2+\cot^{-1}3 \right ]$

$\Rightarrow$         $=\frac{2\pi}{2}-\left [ \tan^{-1}\left ( \frac{1}{2} \right ) +\tan^{-1}\left ( \frac{1}{3} \right )\right ]$

$\Rightarrow$         $=\pi-\left [ \tan^{-1}\left ( \frac{\frac{1}{2}+\frac{1}{3}}{1-\left ( \frac{1}{2} \right )\left ( \frac{1}{3} \right )} \right ) \right ]$                                             $\left [ \because \tan^{-1}a+\tan^{-1}b =\tan^{-1}\left ( \frac{a+b}{1-ab} \right )\right ]$

$\Rightarrow$         $=\pi-\left [ \tan^{-1}\left ( \frac{\frac{5}{6}}{\frac{5}{6}} \right ) \right ]$

$\Rightarrow$         $=\pi-\left [ \tan^{-1}\left ( 1\right ) \right ]$                                                                $\left [ 1=tan^{-1}(tan\frac{\pi}{4}) \right ]$

$\Rightarrow$         $=\pi-\frac{\pi}{4}$

$\Rightarrow$         $=\frac{4\pi-\pi}{4}$

$\Rightarrow$         $=\frac{3\pi}{4}$