Q85 / 225 · JEE Main 2002 · Mathematics

$\cot ^{-1}(\sqrt{\cos \alpha})=\tan ^{-1}(\sqrt{\cos \alpha})=x$, then $\sin x=$

  • Correct answer
    A
    $\tan ^2\left(\frac{\alpha}{2}\right)$
  • B
    $\cot ^2\left(\frac{\alpha}{2}\right)$
  • C
    $\tan \alpha$
  • D
    $\cot \left(\frac{\alpha}{2}\right)$

Solution

$\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$ $\tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x \Rightarrow \tan ^{-1} \frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}=x$ $\Rightarrow \tan ^{-1} \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=x \Rightarrow \tan x=\frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}$ OR $\cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha}$ or $\operatorname{cosec} x=\frac{1+\cos \alpha}{1-\cos \alpha}$ $\sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\frac{1-\left(1-2 \sin ^2 \alpha / 2\right)}{1+2 \cos ^2 \alpha / 2-1}$ or $\sin x=\tan ^2 \frac{\alpha}{2}$

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

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