Find $x$ if $\cot^{-1}\left(\frac{1}{x}\right)+\cos ^{-1}(-x)+\tan^{-1}(x)=\pi$ if $x \lt 0$ Then Find value of  $$\frac{(1-x^2)^{\frac{3}{2}}}{x^2}$$ if
$$ \cot^{-1} \left(\frac{1}{x}\right)+\cos^{-1}(-x)+\tan^{-1}(x)=\pi$$
My try:
Since $x \lt 0$ we have $$ \cot^{-1}\left(\frac{1}{x}\right)=\pi +\tan ^{-1}x$$
Also $$\cos^{-1}(-x)=\pi -\cos ^{-1}x$$
Hence the given equation becomes
$$\pi +2 \tan^{-1} x+\pi -\cos^{-1}{x}=\pi$$
$$\pi+2 \tan^{-1}x=\cos^{-1}x$$
taking $\cos$ both sides we get
$$\frac{x^2-1}{x^2+1}=x$$
Now how to proceed further?

$$\cot^{-1}\left(\frac{1}{x}\right)+\cos ^{-1}(-x)+\tan^{-1}(x)=\pi$$
Suppose $\tan^{-1}(x) = y \implies x = \tan y \implies \cot^{-1}(1/x) = y$
Let $\cos ^{-1}(-x) = z$
$ 2y + z = \pi \implies z = \pi - 2y$
Taking tangents, 
$\tan(\pi - 2y) = \tan z \implies -\tan(2y) = \sin z/\cos z $
We get $-\tan 2y = \frac{\sqrt{1 - (-x)^2}}{-x}$ using $\cos z = -x$
$\frac{2 \tan y}{1 - \tan^2(y)} = \frac{\sqrt{1 - (-x)^2}}{x}$
$\frac{2x}{1-x^2} = \frac{\sqrt{1 - x^2}}{x}$
$\implies \frac{(1-x^2)\sqrt{(1-x^2)}}{x^2} = 2$
