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# Solve the following $tan^{-1} \frac{1-x}{1+x} =\frac{1}{2} tan^{-1} x,(x>0)$

Toolbox:
• $2tan^{-1}y=tan^{-1}\large\frac{2y}{1-y^2}\:\:|y|<1$
Given $tan^{-1} \large\frac{1-x}{1+x} =\large\frac{1}{2} tan^{-1} x,(x>0)$, let's rewrite the given equation as $2tan^{-1}\large\frac{1-x}{1+x}=tan^{-1}x$

We know that $2tan^{-1}y=tan^{-1}\large\frac{2y}{1-y^2}\:\:|y|<1$

By taking $y=\large\frac{1-x}{1+x}$ we get
$2y = \large \frac {2(1-x)}{1+x}$
$1-y^2 = 1 - \large (\frac{1-x}{1+x})^2 = \large\frac{(1+x)^2 - (1-x)^2}{(1+x)^2} = \large\frac{1+2x+x^2-1-x^2+2x}{(1+x)^2} = \large\frac{4x}{(1+x)^2}$
$\large \frac{2y}{1-y^2}= \large \frac {2(1-x)}{1+x} \times \frac{(1+x)^2}{4x} =\large \frac{2(1-x^2)}{4x}$

$\Rightarrow\: 2tan^{-1}\large\frac{1-x}{1+x}=tan^{-1}\large \frac{2(1-x^2)}{4x}$

Using the above formula the given equation becomes $tan^{-1} \large\frac{2(1-x^2)}{4x}=tan^{-1}x$
$\Rightarrow\: \large\frac{2(1-x^2)}{4x}=x$
$\Rightarrow 1-x^2=2x^2$

$\Rightarrow1=3x^2$
$\Rightarrow x=\large\frac{1}{\sqrt 3}$

edited Mar 15, 2013

Toolbox:
• $tan(x-y)=\frac{tanx-tany}{1+tanx.tany}$
• $tan\frac{\pi}{4}=1$
Let $x=tan\theta,\:\Rightarrow\:\theta=tan^{-1}x$
Substituting the value of x in the given equation, we have
$tan^{-1}\bigg(\frac{1-tan\theta}{1+tan\theta}\bigg)=\frac{1}{2}tan^{-1}tan\theta$
We know that $tan\frac{\pi}{4}=1$
$\Rightarrow\:\large\:tan^{-1}\bigg(\large\frac{tan\frac{\pi}{4}-tan\theta}{1+tan\frac{\pi}{4}.tan\theta}\bigg)=\large\frac{1}{2}.\theta$
$\Rightarrow\:\large\:tan^{-1}tan(\frac{\pi}{4}-\theta)=\large\:\frac{1}{2}\theta$
$\Rightarrow\:\large\frac{\pi}{4}-\theta=\large\frac{\theta}{2}$
$\Rightarrow\:\large\frac{3\theta}{2}=\large\frac{\pi}{4}$
$\Rightarrow\:\large\theta=\large\frac{\pi}{6}=\large\:tan^{-1}x$
$\Rightarrow\:\large\:x=\large\:tan\frac{\pi}{6}=\large\frac{1}{\sqrt3}$