Solve: ^ -1 (1-x 1+x )=1 2 ( ^ -1 x ), for x>0. - Vidyadip

Question

Solve: $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\left(\tan ^{-1} x\right)$, for $x>0$.

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Answer

$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\left(\tan ^{-1} x\right)$
$\therefore 2 \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\tan ^{-1} x$
$\therefore \tan ^{-1}\left[\frac{2\left(\frac{1-x}{1+x}\right)}{1-\left(\frac{1-x}{1+x}\right)^2}\right]=\tan ^{-1} x$
$\cdots\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]$
$\therefore \frac{2\left(\frac{1-x}{1+x}\right)(1+x)^2}{(1+x)^2-(1-x)^2}=x$
$\begin{aligned} & \therefore \frac{2(1-x)(1+x)}{\left(1+2 x+x^2\right)-\left(1-2 x+x^2\right)}=x \\ & \therefore \frac{2\left(1-x^2\right)}{1+2 x+x^2-1+2 x-x^2}=x\end{aligned}$
$\begin{aligned} & \therefore \frac{2-2 x^2}{4 x}=x \\ & \therefore 2-2 x^2=4 x^2 \\ & \therefore 6 x^2=2 \quad \therefore x^2=\frac{1}{3} \\ & \therefore x=\frac{1}{\sqrt{3}}\end{aligned}$
$\ldots[\because x>0]$

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