Question

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

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

Answer 1

Answer: (C)

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