Question

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

• A. $\sqrt{ab}$
• B. $\sqrt{2ab}$
• C. $\frac{3\pi }{4}$
• D. $0$

Answer 1

The correct option is A $\sqrt{ab}$

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