Question

The formula ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}x$ holds only for

• A. $x\in R$
• B. $\left|x\right|\le 1$
• C. $x\in (-1,1)$
• D. $x\in (0,\infty )$

Answer 1

The correct option is D $x\in (0,\infty )$

We know that $\cos y\in [-1,1]$

Hence for the above equation to hold,

$-1\le \frac{1-x^2}{1+x^2}\le 1$

$-1\le \frac{1-x^2}{1+x^2}$ and $\frac{1-x^2}{1+x^2}\le 1$

Hence

$-1-x^2\le 1-x^2$ ...(inconclusive).

And

$\frac{1-x^2}{1+x^2}\le 1$

$1-x^2\le 1+x^2$

$0\le 2x^2$

Hence $xϵR$.

However

${\cos }^{-1}(-x)=\pi -x$ where as ${\tan }^{-1}(-x)=-x$

Hence the above equation does not hold good for $xϵR^{-}$

$x<0$

Hence the domain for the above function is

$x>0$

$x\in (0,\infty )$.