Question

If ${\tan }^{-1}\left(\frac{a+x}{a}\right)+{\tan }^{-1}\left(\frac{a-x}{a}\right)=\frac{\mathrm{\pi }}{6}$ then $x^2$ is equal to

• 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

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

Explanation for the correct option

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

$\begin{array}{crclc}\Rightarrow & {\tan }^{-1}\left(\frac{{\displaystyle \frac{a+x}{a}}+{\displaystyle \frac{a-x}{a}}}{1-{\displaystyle \frac{a+x}{a}}\times {\displaystyle \frac{a-x}{a}}}\right)& =& \frac{\mathrm{\pi }}{6}& ;\left[\because {\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(y\right)={\tan }^{-1}\left(\frac{{\displaystyle x+y}}{1-{\displaystyle x}y}\right)\right]\\ \Rightarrow & {\tan }^{-1}\left(\frac{{\displaystyle \frac{2a}{a}}}{{\displaystyle \frac{a^2-\left(a^2-x^2\right)}{a^2}}}\right)& =& \frac{\mathrm{\pi }}{6}& \\ \Rightarrow & {\tan }^{-1}\left(\frac{{\displaystyle 2a^2}}{{\displaystyle x^2}}\right)& =& \frac{\mathrm{\pi }}{6}& \\ \Rightarrow & \frac{{\displaystyle 2a^2}}{{\displaystyle x^2}}& =& \tan \left(\frac{\mathrm{\pi }}{6}\right)& \\ \Rightarrow & \frac{{\displaystyle 2a^2}}{{\displaystyle x^2}}& =& \frac{1}{\sqrt{3}}& \\ \Rightarrow & x^2& =& 2\sqrt{3}{a}^2& \end{array}$

Hence the correct option is option(C) i.e. $2\sqrt{3}{a}^2$