Question

Value of $x$ for which ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}x$ satisfied is $xϵ[a,\infty )$.
Find the value of $a.$

• A. $a=-\infty$
• B. $a=-1$
• C. $a=0$
• D. $a=1$

Answer 1

The correct option is C $a=0$

Given

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

As ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\in [0,\pi ]$

then $2{\tan }^{-1}x\in [0,\pi ]$

${\tan }^{-1}x\in [0,\frac{\pi }{2}]$

$⟹x\in [0,\infty )$

So $a=0$