Question

There exists a positive real number $x$ satisfying $\cos \left({\tan }^{-1}x\right)=x$. The value of ${\cos }^{-1}\left(\frac{x^2}{2}\right)$ is

• A. $\frac{\pi }{10}$
• B. $\frac{\pi }{5}$
• C. $\frac{2\pi }{5}$
• D. $\frac{4\pi }{5}$

Answer 1

The correct option is C $\frac{2\pi }{5}$

$\cos \left({\tan }^{-1}\right(x\left)\right)=x$
$\cos \left({\cos }^{-1}\right(\frac{1}{\sqrt{1+x^2}}\left)\right)=x$
Therefore
$\frac{1}{\sqrt{1+x^2}}=x$
Squaring on both the sides we get
$x^2(x^2+1)=1$
$x^4+x^2-1=0$
Hence $x^2=\frac{-1+\sqrt{5}}{2}$
$\frac{x^2}{2}=\frac{\sqrt{5}-1}{4}$
${\cos }^{-1}\left(\frac{x^2}{2}\right)$ $={\cos }^{-1}\left(\frac{\sqrt{5}-1}{4}\right)$
$={72}^0$
$=\frac{2\pi }{5}$