Question

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

Answer 1

Here, for domain $\left|\frac{1+x^2}{1-x^2}\right|\le 1$
$\Rightarrow |1-x^2|\le 1+x^2$
$\because 1+x^2>0,$ for all $x\in R$
which is true for all $x$ as $1+x^2>1-x^2$
$\therefore x\in R$
For Range: $y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
$\Rightarrow y\in (0,\pi )$
Define the curve: Let $x=\tan \theta$
$\therefore y={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$
$={\cos }^{-1}\left(\cos 2\theta \right)$
$=\{\begin{array}{c}2\theta ;2\theta \ge 0\\ -2\theta ;2\theta <0\end{array}$
Since $\tan \theta =x\Rightarrow \theta ={\tan }^{-1}x$
$\Rightarrow y=\{\begin{array}{c}2{\tan }^{-1}x;{\tan }^{-1}x\ge 0\\ -2{\tan }^{-1}x;{\tan }^{-1}x<0\end{array}$
So, the graph of
$y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
$=\{\begin{array}{c}2{\tan }^{-1}x;{\tan }^{-1}x\ge 0\\ -2{\tan }^{-1}x;{\tan }^{-1}x<0\end{array}$ is as shown
Thus, the graph for
$y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
$=\{\begin{array}{c}2{\tan }^{-1}x;x\ge 0\\ -2{\tan }^{-1}x;x<0\end{array}$