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. $2ab$
• D. $ab$

Answer 1

Answer: (A)

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