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