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. $ab$
• B. $\frac{a}{b}$
• C. $\frac{b}{a}$
• D. $\sqrt{ab}$

Answer 1

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

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