Q43 / 120 · VITEEE 2008 · Mathematics

The simplified expression of $\sin \left(\tan ^{-1} x\right)$, for any real number $x$ is given by

  • A
    $\frac{1}{\sqrt{1+x^{2}}}$
  • Correct answer
    B
    $\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}$
  • C
    $-\frac{1}{\sqrt{1+x^{2}}}$
  • D
    $-\frac{x}{\sqrt{1+x^{2}}}$

Solution

Let $\tan ^{-1} x=\theta$ $\Rightarrow x=\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$ $\Rightarrow x=\sqrt{1-\sin ^{2} \theta}=\sin \theta$ $\Rightarrow x^{2}\left(1-\sin ^{2} \theta\right)=\sin ^{2} \theta$ $\Rightarrow x^{2}=\sin ^{2} \theta\left(1+x^{2}\right)$ $\Rightarrow \sin ^{2} \theta=\frac{x^{2}}{1+x^{2}} \Rightarrow \sin \theta=\frac{x}{\sqrt{1+x^{2}}}$ $\Rightarrow \theta=\sin ^{-1} \frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}$ $\begin{aligned} \Rightarrow \tan ^{-1} \mathrm{x} &=\sin ^{-1} \frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}} \\ \text { Now, } \sin \left(\tan ^{-1} \mathrm{x}\right) &=\sin \left(\sin ^{-1} \frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}\right) \\ &=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}} \end{aligned}$

Hence, the correct answer is option: 'B'.

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