Question

$y=f\left(x\right)={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\{\begin{array}{c}2{\tan }^{-1}x\text{if}x\ge 0\\ -2{\tan }^{-1}x\text{if}x<0\end{array}$

• A. True
• B. False

Answer 1

The correct option is A True

Let $x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x$

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

$\Rightarrow \cos y=\left|\frac{1-x^2}{1+x^2}\right|$

$\Rightarrow \cos y=\left|\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right|$

$\Rightarrow \cos y=\left|\frac{1-\frac{{\sin }^2\theta }{{\cos }^2\theta }}{{\sec }^2\theta }\right|$

$\Rightarrow \cos y=\left|\frac{\frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta }}{\frac{1}{{\cos }^2\theta }}\right|$

$\Rightarrow \cos y=|{\cos }^2\theta -{\sin }^2\theta |$

$\Rightarrow \cos y=\left|\cos 2\theta \right|$

$\Rightarrow y=\left|2\theta \right|$

$\Rightarrow y=\left|2{\tan }^{-1}x\right|$

$\therefore y=\{\begin{array}{c}2{\tan }^{-1}x;ifx\ge 0\\ -2{\tan }^{-1}x;ifx<0\end{array}$