Q. The simplified expression of $\sin{(\tan^{- 1}x)}$, 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) $- \frac{x}{\sqrt{1 + x^{2}}}$
Explanation
(b)
Let $\tan^{- 1}{x = \theta}$
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$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tan{\theta = x}$
$\therefore\ \sin{\left( \tan^{- 1}x \right) = \sin\theta}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{BC}{AC} = \frac{x}{\sqrt{1 + x^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.