Question

Simplify $\sin {\cot }^{-1}\tan {\cos }^{-1}x$.

• A. $\pi -x$
• B. $\pi +x$
• C. $x$
• D. $-x$

Answer 1

The correct option is C $x$

Let,

${\cos }^{-1}x=\theta$
$\Rightarrow$ $x=\cos \theta$
$\sec \theta =\frac{1}{x}$
$\Rightarrow$ $\tan \theta =\sqrt{{\sec }^2\theta -1}=\sqrt{\frac{1}{x^2}-1}=\frac{1}{\left|x\right|}\sqrt{1-x^2}$

Now, $\sin {\cot }^{-1}\tan {\cos }^{-1}x=\sin {\cot }^{-1}\left(\frac{1}{\left|x\right|}\sqrt{1-x^2}\right)$

Again, putting $x=\sin \theta$, we get
$\sin {\cot }^{-1}\left(\frac{1}{\left|x\right|}\sqrt{1-x^2}\right)=\sin {\cot }^{-1}\left(\frac{\sqrt{1-{\sin }^2\theta }}{\left|\sin \theta \right|}\right)=\sin {\cot }^{-1}\left|\cot \theta \right|=\sin \theta =x$