Question

If $\frac{1}{\sqrt{2}}<x<1$ then ${\cos }^{-1}x+{\cos }^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)$ is equal to

• A. $2{\cos }^{-1}x-\frac{\pi }{4}$
• B. $2{\cos }^{-1}x$
• C. $\frac{\pi }{4}$
• D. 0

Answer 1

Answer: (C)

$\frac{\pi }{4}$

$co{s}^{-1}\left(x\right)+co{s}^{-1}(\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}\sqrt{1-x^2})$
$=co{s}^{-1}\left(x\right)+co{s}^{-1}\left(\frac{1}{\sqrt{2}}\right)-co{s}^{-1}\left(x\right)$
$=co{s}^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$=\frac{\pi }{4}$