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#### Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 5

$\frac{-\pi}{2}$

Hint:

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

Given:

$x< 0,\tan^{-1}x+\tan^{-1}\frac{1}{x}$

Solution:

Using the property of inverse trigonometry

$\begin{gathered} \tan ^{-1} x+\tan ^{-1} \frac{1}{x}=\tan ^{-1}\left(\frac{x+\frac{1}{x}}{1-1}\right)\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan ^{-1} a+\tan ^{-1} b=\tan ^{-1}\left(\frac{a+b}{1-a b}\right)\right] \\ \end{gathered}$

$=\tan ^{-1}\left(\frac{\frac{x^{2}+1}{x}}{0}\right)$

Since,$\left ( x+\frac{1}{x} \right )$=Negative value $=$ integer $= -a$

$\tan^{-1}\left ( \frac{-a}{0} \right )$

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

Using value of inverse trigonometry,

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

Substituting the value, we get

$\Rightarrow \tan^{-1}x+\tan^{-1}\frac{1}{x}=\frac{-\pi}{2}$