Q72 / 180 · COMEDK 2018 · Mathematics

The value of $\sin \left[\cot ^{-}\left\{\cos \left(\tan ^{-1} x\right)\right\}\right]$ is

  • Correct answer
    A
    $\left(\sqrt{\frac{1+x^{2}}{2+x^{2}}}\right)$
  • B
    $\left(\sqrt{\frac{2+x^{2}}{1+x^{2}}}\right)$
  • C
    $\left(\sqrt{\frac{x^{2}-2}{x^{2}-1}}\right)$
  • D
    $\left(\sqrt{\frac{x^{2}-1}{x^{2}-2}}\right)$

Solution

We have $\sin \left[\cot ^{-1}\left\{\cos \left(\tan ^{-1} x\right)\right\}\right]$ Now, $\cos \left(\tan ^{-1} x\right)=\cos \left[\cos ^{-1} \frac{1}{\sqrt{x^{2}+1}}\right]$ $=\frac{1}{\sqrt{x^{2}+1}}$ $\left[\because \cos \left(\cos ^{-1} x\right)=x, \forall x \in[-1,1]\right]$ So, $\cot ^{-1}\left(\frac{1}{\sqrt{x^{2}+1}}\right)=\sin ^{-1}\left(\frac{\sqrt{x^{2}+1}}{\sqrt{x^{2}+2}}\right)$ Thus, $\sin \left[\cot ^{-1}\left(\frac{1}{\sqrt{x^{2}+1}}\right)\right]=\sin \left[\sin ^{-1}\left(\frac{\sqrt{x^{2}+1}}{\sqrt{x^{2}+2}}\right)\right]$ $=\frac{\sqrt{\left(x^{2}+1\right)}}{\sqrt{x^{2}+2}} \quad\left[\because \sin \left(\sin ^{-1} x\right)=x, \forall x \in[-1,1]\right.$

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

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