Question

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

Answer 1

We know

​${\tan }^{-1}x={\cot }^{-1}\frac{1}{x}$

$\therefore {\tan }^{-1}\frac{b}{x}={\cot }^{-1}\frac{1}{\left({\displaystyle \frac{\mathit{b}}{\mathit{x}}}\right)}={\cot }^{-1}\frac{\mathit{x}}{\mathit{b}}$ .....(1)
So,

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

$\Rightarrow {\tan }^{-1}\frac{a}{x}+{\cot }^{-1}\frac{x}{b}=\frac{\mathrm{\pi }}{2}$ [Using (1)]

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

$\Rightarrow x^2=ab$

$\Rightarrow x=\sqrt{ab}$

If tan⁻¹$\frac{a}{x}$+ tan⁻¹$\frac{b}{x}=\frac{\mathrm{\pi }}{2}$, then x = $\underline{\sqrt{ab}}$.