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

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

Inverse Trigonometric Functions

Solution:

Let $\tan ^{-1}\left(\frac{ a }{ x }\right)+\tan ^{-1}\left(\frac{ b }{ x }\right)=\frac{\pi}{2}$
$\Rightarrow \tan ^{-1}\left(\frac{\frac{a}{x}+\frac{b}{x}}{1-\frac{a b}{x^{2}}}\right)=\frac{\pi}{2} $
$\Rightarrow \frac{\frac{a}{x}+\frac{b}{x}}{1-\frac{a b}{x^{2}}}=\tan \frac{\pi}{2}$
$\Rightarrow 1-\frac{ ab }{ x ^{2}}=0 $
$\Rightarrow x ^{2}= ab $
$\Rightarrow x =\sqrt{ ab }$