If $\tan ^{-1} \frac{a}{x}+\tan ^{-1} \frac{b}{x}=\frac{\pi}{2}$, then $x=$ _________________.
We know
$\tan ^{-1} x=\cot ^{-1} \frac{1}{x}$
$\therefore \tan ^{-1} \frac{b}{x}=\cot ^{-1} \frac{1}{\left(\frac{b}{x}\right)}=\cot ^{-1} \frac{x}{b}$ ....(1)
So,
$\tan ^{-1} \frac{a}{x}+\tan ^{-1} \frac{b}{x}=\frac{\pi}{2}$ (Given)
$\Rightarrow \tan ^{-1} \frac{a}{x}+\cot ^{-1} \frac{x}{b}=\frac{\pi}{2}$ [Using (1)]
$\Rightarrow \frac{a}{x}=\frac{x}{b}$ $\left(\tan ^{-1} y+\cot ^{-1} y=\frac{\pi}{2}\right)$
$\Rightarrow x^{2}=a b$
$\Rightarrow x=\sqrt{a b}$
If $\tan ^{-1} \frac{a}{x}+\tan ^{-1} \frac{b}{x}=\frac{\pi}{2}$, then $x=\sqrt{a b}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.