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Question:

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

Solution:

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 $x=\sqrt{a b}$

 

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