question_answer

The simplified expression of \[\sin \,(ta{{n}^{-1}}x),\] for any real number x is given by

A)  \[\frac{1}{\sqrt{1+{{x}^{2}}}}\]

B)  \[\frac{x}{\sqrt{1+{{x}^{2}}}}\]

C)  \[-\frac{1}{\sqrt{1+{{x}^{2}}}}\]

D)  \[-\frac{x}{\sqrt{1+{{x}^{2}}}}\]

Correct Answer: B

Solution :

Let \[{{\tan }^{-1}}x=\theta \] \[\Rightarrow \] \[\tan \theta =x\] \[\because \]\[\sin \,({{\tan }^{-1}}x)=\sin \theta \] \[=\frac{BC}{AC}=\frac{x}{\sqrt{1+{{x}^{2}}}}\]