MCQ
The value of $\sin \left[\cot ^{-1}\left(\tan \cos ^{-1} x\right)\right]$ is equal to
- ✓$x$
- B$\frac{\pi}{2}$
- C$1$
- D$\pi$
✓
Answer
Correct option: A.
$x$
(A) Let $\cos ^{-1} x=\theta \Rightarrow x=\cos \theta \Rightarrow \sec \theta=\frac{1}{x}$
$\Rightarrow \tan \theta=\sqrt{\sec ^2 \theta-1}=\sqrt{\frac{1}{x^2}-1}=\frac{1}{x} \sqrt{1-x^2}$
Now,
$\sin \left[\cot ^{-1}(\tan \theta)\right]=\sin \left[\cot ^{-1}\left(\frac{1}{x} \sqrt{1-x^2}\right)\right]$
Again, putting $x=\sin \theta$
$\therefore \quad \sin \cot ^{-1}\left(\frac{1}{x} \sqrt{1-x^2}\right)=\sin \cot ^{-1}\left(\frac{\sqrt{1-\sin ^2 \theta}}{\sin \theta}\right)$
$\begin{array}{l}=\sin \cot ^{-1}(\cot \theta) \\ =\sin \theta=x\end{array}$
$\Rightarrow \tan \theta=\sqrt{\sec ^2 \theta-1}=\sqrt{\frac{1}{x^2}-1}=\frac{1}{x} \sqrt{1-x^2}$
Now,
$\sin \left[\cot ^{-1}(\tan \theta)\right]=\sin \left[\cot ^{-1}\left(\frac{1}{x} \sqrt{1-x^2}\right)\right]$
Again, putting $x=\sin \theta$
$\therefore \quad \sin \cot ^{-1}\left(\frac{1}{x} \sqrt{1-x^2}\right)=\sin \cot ^{-1}\left(\frac{\sqrt{1-\sin ^2 \theta}}{\sin \theta}\right)$
$\begin{array}{l}=\sin \cot ^{-1}(\cot \theta) \\ =\sin \theta=x\end{array}$
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