Question

If ${\tan }^{-1}\frac{a+x}{a}+{\tan }^{-1}\frac{a-x}{a}=\frac{\pi }{6}$, then $x^2$ is

• A. $2\sqrt{3}a$
• B. $\sqrt{3}a$
• C. $2\sqrt{3}{a}^2$
• D. None of these

Answer 1

Answer: (C)

Given, ${\tan }^{-1}\frac{a+x}{a}+{\tan }^{-1}\frac{a-x}{a}=\frac{\pi }{6}$
$\Rightarrow {\tan }^{-1}\left[\frac{\frac{a+x}{a}+\frac{a-x}{a}}{1-\left(\frac{a+x}{a}\right)\left(\frac{a-x}{a}\right)}\right]=\frac{\pi }{6}$
$\Rightarrow {\tan }^{-1}\left[\frac{\frac{a+x+a-x}{a}}{\frac{a^2-a^2+x^2}{a^2}}\right]=\frac{\pi }{6}$
$\Rightarrow \frac{2a^2}{x^2}=\tan \frac{\pi }{6}$
$\Rightarrow \frac{2a^2}{x^2}=\frac{1}{\sqrt{3}}$
$\Rightarrow x^2=2\sqrt{3}{a}^2$