Question

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. $-\frac{x}{\sqrt{1+x^{2}}}$

Answer 1

Answer: (B)

Let tan$^{-1}$ x = $\theta$
$\Rightarrow x = tan \theta = \frac{sin \theta }{cos \theta }=\frac{sin \theta }{\sqrt{1-sin^{2} \theta }}$
$\Rightarrow x =\sqrt{1-sin^{2} \theta }=sin \theta$
$\Rightarrow x^{2}\left(1-sin^{2} \theta \right)=sin^{2} \theta $
$\Rightarrow x^{2}=sin^{2} \theta \left(1+x^{2}\right)$
$\Rightarrow sin^{2} \theta =\frac{x^{2}}{1+x^{2}} \Rightarrow sin \theta =\frac{x}{\sqrt{1+x^{2}}}$
$\Rightarrow \theta =sin^{-1} \frac{x}{\sqrt{1+x^{2}}}$
$\Rightarrow tan^{-1} x =sin^{-1} \frac{x}{\sqrt{1+x^{2}}}$
Now, sin $\left(tan^{-1}x\right) = sin\left( sin^{-1} \frac{x}{\sqrt{1+x^{2}}}\right)$
$\quad\quad\quad\quad\quad\quad=\frac{x}{\sqrt{1+x^{2}}}$