Question

If ${\tan }^{-1}\left(\frac{a}{x}\right)+{\tan }^{-1}\left(\frac{b}{x}\right)=\frac{\pi }{2}$, then $x=$

• A. $\sqrt{ab}$
• B. $\sqrt{2ab}$
• C. $2ab$
• D. $ab$

Answer 1

The correct option is A

$\sqrt{ab}$

Explanation for the correct option:

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

$\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x+y}}{{\displaystyle 1-xy}}\right)$

Using the above formula in given equation we get;

$\begin{array}{c}\Rightarrow {\tan }^{-1}\left(\frac{a}{x}\right)+{\tan }^{-1}\left(\frac{b}{x}\right)=\frac{\pi }{2}\\ \Rightarrow \frac{\left(\frac{a}{x}+\frac{b}{x}\right)}{\left(1-\left(\frac{ab}{x^2}\right)\right)}=\tan \frac{\pi }{2}=0\\ \Rightarrow 1-\frac{ab}{x^2}=0\\ \Rightarrow x^2=ab\\ \Rightarrow x=\sqrt{ab}\end{array}$

Hence option (A) is correct.