15. \sin(\tan^{-1}x),\;|x|<1  is equal to 

        (A)    \frac{x}{\sqrt{1-x^2}}

        (B)    \frac{1}{\sqrt{1-x^2}}

        (C)    \frac{1}{\sqrt{1+x^2}}

        (D)    \frac{x}{\sqrt{1+x^2}}

Answers (1)

Let \tan^{-1}x = y then we have;

\tan y = x   or

y=\sin^{-1} \left ( \frac{x}{\sqrt{1+x^2}} \right ) \Rightarrow \tan^{-1} x = \sin^{-1} \left ( \frac{x}{\sqrt{1+x^2}} \right)

\Rightarrow \sin \left ( \tan^{-1} x \right ) = \sin\left ( \sin^{-1}\left ( \frac{x}{\sqrt{1+x^2}} \right ) \right ) = \frac{x}{\sqrt {1+x^2}}

Hence the correct answer is D.

 

 

 

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