Q

# Solve the following equations: 14.

Solve the following equations:

14. $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x,\;(x>0)$

Views

Given equation is

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

L.H.S can be written as;

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

Using the formula $\left [ \tan^{-1}x -\tan^{-1}y = \tan^{-1} \frac{x-y}{1+xy} \right ]$

So, we have $\tan^{-1}1 - \tan^{-1}x = \frac{1}{2} \tan^{-1}x$

$\Rightarrow \tan^{-1}1= \frac{3}{2} \tan^{-1}x$

$\Rightarrow\frac{\pi}{4}= \frac{3}{2} \tan^{-1}x$

$\Rightarrow \tan^{-1}x = \frac{\pi}{6}$

$\Rightarrow x= \tan \frac{\pi}{6} = \frac{1}{\sqrt3}$

Hence the value of $x= \frac{1}{\sqrt3}$.

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