Question

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

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

Answer 1

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

Formula ${\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2})x,y>0$

$\therefore {\cos }^{-1}x+{\cos }^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)={\cos }^{-1}[\frac{x(x+\sqrt{1-x^2})}{\sqrt{2}}-\sqrt{1-x^2}\left(\sqrt{1-\frac{1+2x\sqrt{1-x^2}}{2}}\right)]$

$={\cos }^{-1}(\frac{x(x+\sqrt{1-x^2})}{\sqrt{2}}-\sqrt{1-x^2}\sqrt{\frac{1-2x\sqrt{1-x^2}}{2}})$

$={\cos }^{-1}(\frac{x(x+\sqrt{1-x^2})}{\sqrt{2}}-\frac{\sqrt{1-x^2}}{\sqrt{2}}(x-\sqrt{1-x^2}\left)\right)$

$={\cos }^{-1}(\frac{x^2+x\sqrt{1-x^2}}{\sqrt{2}}-\frac{(x-\sqrt{1-x^2}\sqrt{1-x^2})}{\sqrt{2}})$

$={\cos }^{-1}(\frac{x^2+x\sqrt{1-x^2}}{\sqrt{2}}-\left(\frac{x\sqrt{1-x^2}+x^2-1}{\sqrt{2}}\right))$

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

$=\frac{\pi }{4}$