If x < 0, then write the value

Question:

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

Solution:

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-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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