Question

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

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

Answer 1

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

Let ${\tan }^{-1}\left(x\right)=\theta$
$\Rightarrow \tan \theta =x$

By right angled triangle,
$\cos \theta =\frac{1}{\sqrt{1+x^2}}$
$\Rightarrow \frac{1}{\sqrt{1+x^2}}=x$
$\Rightarrow x^2(1+x^2)=1\Rightarrow x^4+x^2-1=0\Rightarrow x^2=\frac{-1\pm \sqrt{5}}{2}$
But $x^2$ cannot be negative.
$x^2=\frac{\sqrt{5}-1}{2}$
$\Rightarrow \frac{x^2}{2}=\frac{\sqrt{5}-1}{4}$

Now, ${\cos }^{-1}\left(\frac{\sqrt{5}-1}{4}\right)={\cos }^{-1}\left(\sin \frac{\pi }{10}\right)$
$={\cos }^{-1}\left(\cos \frac{2\pi }{5}\right)=\frac{2\pi }{5}(\because {\cos }^{-1}(\cos x)=x,x\in [0,\pi \left]\right)$