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

$\sqrt{ab}$

$\tan^{-1}\frac{a}{x}+\tan^{-1}\frac{b}{x}=\frac{\pi}{2}$

So $\tan^{-1}\left(\frac{\frac{a}{x}+\frac{b}{x}}{1-\frac{ab}{x^2}}\right)=\frac{\pi}{2}$  as $\left\{\tan^{-1}x+\tan^{-1}y=\tan^{-1}(\frac{x+y}{1-xy})\right\}$

So $1 -\frac{ab}{x^2}= 0$ as $\left(\underset{y→0}{\lim}\tan^{-1}(\frac{x}{y})=\frac{\pi}{2}\right)$

So $1=\frac{ab}{x^2}$

$ab=x^2⇒ x = \sqrt{ab}$