Question

If $x>1$, then $2{\tan }^{-1}x$ cannot be equal to

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

Answer 1

The correct option is D $\pi -{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

Let $\alpha ={\tan }^{-1}x$

$x=\tan \alpha$

$\therefore \cos 2\alpha =2{\cos }^2\alpha -1$

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

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

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

Put $x=tan\theta$ for all other options and check.