Question

If x < 0, then write the value of cos⁻¹$\left(\frac{1-x^2}{1+x^2}\right)$ in terms of tan⁻¹ x.

Answer 1

Let $x=\tan y$
Then,
$$\begin{gathered} {\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)={\cos }^{-1}\left(\frac{1-{\tan }^{2}y}{1+{\tan }^{2}y}\right) \\ ={\cos }^{-1}\left(\cos 2y\right)\left[\because \frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}=\cos 2x\right] \\ =2y...\left(1\right) \end{gathered}$$

The value of x is negative.
So, let x = $-a$ where a > 0.

$$\begin{gathered} -a=\tan y \\ \Rightarrow y={\tan }^{-1}\left(-a\right) \end{gathered}$$
Now,

$$\begin{gathered} {\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2y\left[\mathrm{Using}\left(1\right)\right] \\ =2{\tan }^{-1}\left(-a\right) \\ =-2{\tan }^{-1}x\left[\because x=-a\right] \end{gathered}$$