Question

If $\frac{1}{\sqrt{2}}<x<1$, then find the value of ${\cos }^{-1}x+{\cos }^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)$.

• A. $-\frac{\pi }{4}$
• B. $+\frac{\pi }{4}$
• C. $-\frac{\pi }{2}$
• D. $+\frac{\pi }{2}$

Answer 1

The correct option is B $+\frac{\pi }{4}$

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

Let ${\cos }^{-1}x=\theta$ or $x=\cos \theta$

For $\frac{1}{\sqrt{2}}<x<1$, $0<\theta <\frac{\pi }{4}$

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

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

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

$=\theta +{\cos }^{-1}\left(\cos (\theta -\frac{\pi }{4})\right)$

Now,
$(\theta -\frac{\pi }{4})\in (-\frac{\pi }{4},0)$, which is not the principle values of ${\cos }^{-1}$ function.

But $(\frac{\pi }{4}-\theta )\in (0,\frac{\pi }{4})$

$\Rightarrow$ $\theta +{\cos }^{-1}\left(\cos (\theta -\frac{\pi }{4})\right)$

$=\theta +{\cos }^{-1}\left(\cos (\frac{\pi }{4}-\theta )\right)$

$=\theta +{\cos }^{-1}(\frac{\pi }{4}-\theta )$

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