Question

For $0\le {\cos }^{-1}x\le \pi$ and $-\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}$, the value of $\cos ({\sin }^{-1}x+2{\cos }^{-1}x)$ at $x=\frac{1}{5}$ is:

• A. $\frac{-2\sqrt{6}}{5}$
• B. $\frac{\sqrt{6}}{5}$
• C. $\frac{-2\sqrt{5}}{7}$
• D. $\frac{-\sqrt{5}}{4}$

Answer 1

The correct option is A $\frac{-2\sqrt{6}}{5}$

Since, ${\sin }^{-1}\theta +{\cos }^{-1}\theta =\frac{\pi }{2}$
$\therefore \cos ({\sin }^{-1}x+2{\cos }^{-1}x)=\cos (\frac{\pi }{2}+{\cos }^{-1}x)=-\cos \left({\sin }^{-1}x\right)=-\cos \left({\cos }^{-1}\sqrt{1-x^2}\right)$
$=-\cos \left({\cos }^{-1}\sqrt{1-{\left(\frac{1}{5}\right)}^2}\right)$

$=-\cos \left({\cos }^{-1}\sqrt{\frac{24}{25}}\right)$
$=-\cos \left({\cos }^{-1}\frac{2\sqrt{6}}{5}\right)=\frac{-2\sqrt{6}}{5}$