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)

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