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

Answer 1

The correct option is C $\frac{\pi }{4}$

$\frac{1}{\sqrt{2}}<x<1$
${\cos }^{-1}x+{\cos }^{-1}\left[\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right]$
$={\cos }^{-1}x+{\cos }^{-1}[x.\frac{1}{\sqrt{2}}+\sqrt{1-x^2}.\sqrt{1-\frac{1}{2}}]$
Let $y=\frac{1}{\sqrt{2}}$
$={\cos }^{-1}x+{\cos }^{-1}[xy+\sqrt{1-x^2}\sqrt{1-y^2}]$
$={\cos }^{-1}x+[{\cos }^{-1}y-{\cos }^{-1}x]$
${\cos }^{-1}x+{\cos }^{-1}\frac{1}{\sqrt{2}}-{\cos }^{-1}x$
$={\cos }^{-1}\frac{1}{\sqrt{2}}$
$=\frac{\pi }{4}$