Browse Questions

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

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 $tan\frac{\pi}{4}=1$
$\Rightarrow\:\large\:tan^{-1}\bigg(\large\frac{tan\frac{\pi}{4}-tan\theta}{1+tan\frac{\pi}{4}.tan\theta}\bigg)=\large\frac{1}{2}.\theta$
$\Rightarrow\:\large\:tan^{-1}tan(\frac{\pi}{4}-\theta)=\large\:\frac{1}{2}\theta$
$\Rightarrow\:\large\frac{\pi}{4}-\theta=\large\frac{\theta}{2}$
$\Rightarrow\:\large\frac{3\theta}{2}=\large\frac{\pi}{4}$
$\Rightarrow\:\large\theta=\large\frac{\pi}{6}=\large\:tan^{-1}x$
$\Rightarrow\:\large\:x=\large\:tan\frac{\pi}{6}=\large\frac{1}{\sqrt3}$
edited Mar 21, 2013