Question

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

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

Answer 1

The correct option is C $2\sqrt{3}{a}^2$

${\tan }^{-1}\left(\frac{a+x}{a}\right)+{\tan }^{-1}\left(\frac{a-x}{a}\right)=\frac{\pi }{6}$

${\tan }^{-1}\frac{(\frac{a+x}{a}+\frac{a-x}{a})}{(1-\left(\frac{a+x}{a}\right)\left(\frac{a-x}{a}\right))}=\frac{\pi }{6}$

${\tan }^{-1}\left(\frac{\frac{2a}{a}}{1\frac{1a^2-x^2}{a^2}}\right)=\frac{\pi }{6}$

${\tan }^{-1}\left(\frac{2a^2}{a^2-a^2+x^2}\right)=\frac{\pi }{6}$

$\frac{2a^2}{x^2}=\tan \frac{\pi }{6}$

$x^2=\frac{2a^2}{\frac{1}{\sqrt{3}}}$

$x^2=2\sqrt{3}{a}^2$