MCQ
Solve $\sin \left(\tan ^{-1} x\right),|x|<1$ is equal to
- A$\frac{1}{\sqrt{1+x^{2}}}$
- B$\frac{1}{\sqrt{1-x^{2}}}$
- ✓$\frac{x}{\sqrt{1+x^{2}}}$
- D$\frac{x}{\sqrt{1-x^{2}}}$
Let $\tan ^{-1} x=y .$ Then
$y=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)$ $\Rightarrow \tan ^{-1} x=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)$
$\Rightarrow \sin \left(\tan ^{-1} x\right)=\sin \left(\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)\right)=\frac{x}{\sqrt{1+x^{2}}}$
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