If $x<0$, then write the value of $\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ in terms of $\tan ^{-1} x$
Let $x=\tan y$
Then,
$\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\cos ^{-1}\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right)$
$=\cos ^{-1}(\cos 2 y) \quad\left[\because \frac{1-\tan ^{2} x}{1+\tan ^{2} x}=\cos 2 x\right]$
$=2 y \quad \ldots(1)$
The value of x is negative.
So, let x = −a where a > 0.
$-a=\tan y$
$\Rightarrow y=\tan ^{-1}(-a)$
Now,
$\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 y$ [Using (1)]
$=2 \tan ^{-1}(-a)$
$=-2 \tan ^{-1} x$ $[\because x=-a]$
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