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  4. If tan -1 (a+x/a)+ tan -1 (a-x/a)=(π/6), then x2 is

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

Inverse Trigonometric Functions

Solution:

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