Question
$\sin (\tan^{-1} x), |x| < 1$ का मान बराबर है:

Answer

$(\sin \tan^{-1} x)$
$= \sin [\sin^{-1} \frac{x}{\sqrt{1+x^{2}}}] (\because \tan^{-1} x = \sin^{-1} \frac{x}{\sqrt{1+x^{2}}})$
$= \frac{x}{\sqrt{1+x^{2}}}$

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