Question

If ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{2}$ , then the value of $x$ is equal to :

• A. $\frac{1}{\sqrt{6}}$
• B. $\frac{1}{6}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{2}}$
• E. $\frac{1}{3}$

Answer 1

Answer: (A)

$\frac{1}{\sqrt{6}}$

Given, ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{2}$
Thus ${\tan }^{-1}\left(\frac{2x+3x}{1-2x\times 3x}\right)=\frac{\pi }{2}$
$\Rightarrow \frac{5x}{1-6x^2}=\tan \frac{\pi }{2}=\frac{1}{0}$
$\Rightarrow 1-6x^2=0$
$\Rightarrow 6x^2=1$
$\Rightarrow x=\pm \frac{1}{\sqrt{6}}$
${\Rightarrow }^{\prime }x=\frac{1}{\sqrt{6}}$
$\left[\begin{array}{c}\because x=-\frac{1}{\sqrt{6}}\\ \text{does not satisfy the given equation}\end{array}\right]$