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  4. If tan-1((a/x))+ tan-1((b/x))=(π/2), then x=

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

Inverse Trigonometric Functions

Solution:

$tan^{-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$
$\Rightarrow x=\sqrt{ab}$