Q

# sin ( tan inverse x), |x| < 1 is equal to (A) (B) (C) (D)

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)    $\frac{x}{\sqrt{1+x^2}}$

Views

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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