Question

$\sin \left[{\cot }^{-1}\left\{\tan \left({\cos }^{-1}x\right)\right\}\right]$ is equal to

• A. $x$
• B. $\sqrt{1-x^2}$
• C. $\frac{1}{x}$
• D. None of these

Answer 1

The correct option is A $x$

given, $\sin \left[{\cot }^{-1}\left\{\tan \left({cos}^{-1}x\right)\right\}\right]$

Let ${cos}^{-1}x=a$

$\Rightarrow \cos a=x$

$\Rightarrow \sin a=\sqrt{1-{\cos }^{2}a}=\sqrt{1-x^2}$

$\Rightarrow \tan a=\frac{\sin a}{\cos a}=\frac{\sqrt{1-x^2}}{x}$

$\Rightarrow a={\tan }^{-1}\frac{\sqrt{1-x^2}}{x}$

$\therefore a={cos}^{-1}x={\tan }^{-1}\frac{\sqrt{1-x^2}}{x}$

$\therefore \sin \left[{\cot }^{-1}\left\{\tan \left({\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\right)\right\}\right]$

$=\sin \left[{\cot }^{-1}\left\{\frac{\sqrt{1-x^2}}{x}\right\}\right]$

Let ${\cot }^{-1}\left\{\frac{\sqrt{1-x^2}}{x}\right\}=b$

$\Rightarrow \cot b=\frac{\sqrt{1-x^2}}{x}$

$\Rightarrow {cosec}^{2}b=1+{\cot }^{2}b=1+\frac{1-x^2}{x^2}=\frac{x^2+1-x^2}{x^2}=\frac{1}{x^2}$

$\Rightarrow cosecb=\frac{1}{x}$

$\Rightarrow \sin b=\frac{1}{cosecb}=\frac{1}{\frac{1}{x}}=x$

$\Rightarrow b={\sin }^{-1}x$

$=\sin \left[b\right]$

$=\sin \left[{\sin }^{-1}x\right]=x$