Solve the following equations:

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

Answers (1)

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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