$y''$ in terms of $y$ if $y=\cos^{-1}x$ 
Find $\frac{d^2y}{dx^2}$ if $y=\cos^{-1}x$ in terms of $y$ only.

The solution is given as $y''=-\cot y.\csc^2y$, but is it the only solution ?
My Attempt
$$
y'=\frac{-1}{\sqrt{1-x^2}}
\\y''=\frac{1}{2\sqrt{1-x^2}.(1-x^2)}.-2x=\frac{-x}{(1-x^2)^{{3}/{2}}}
$$
Here $y=\cos^{-1}x\implies x=\cos y$
$$
y''=\frac{-\cos y}{(\sin^2y)^{3/2}}=\frac{-\cos y}{|\sin y|^3}=\frac{-\cos y}{|\sin y|}.\frac{1}{\sin^2y}=\color{blue}{\pm\csc^2y.\cot y}
$$
 A: I would do it this way:
$$x=\cos(y)$$
Differentiate both sides with respect to $x$:
$$1=-\sin(y)\frac{\mathrm{d}y}{\mathrm{d}x}$$
$$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sin(y)}$$
Now differentiate again:
$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}y}{\mathrm{d}x}\cot(y)\csc(y)$$
$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\frac{1}{\sin(y)}\cot(y)\csc(y)$$
$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\frac{\cos(y)}{\sin^3(y)}$$
And @Andreas covered your sign error already, but a little bit of extension:  
Osberve that $\text{ran}{(\cos^{-1})}=[0,\pi]$, and the $\sin(x)\geq 0$ iniqualits holds $\forall x \in [0, \pi]$, so you need to pick the positive root. But you can avoid it with using the method I've used.
Note: If you invert the $\cos$ on the interval $[-\pi, 0]$, then the $\sin$ will be negative (or 0) there, so you will need to choose the negative sine. But the usual way of inverting the cosine is on the $[0, \pi]$ interval.
A: You are introducing an option of signs which does not exist. Start with your previous result here:
$$
y''=\frac{- x}{\sqrt{1-x^2}.(1-x^2)}
$$
Observe that the root is only the positive root (!) which is $+ \sin y$, so you get 
$$
y''=\frac{-\cos(y)}{\sin^3(y)} = - \csc^2y.\cot y
$$
This is the only solution (apart from the usual angle shift properties of the trig functions).
A: $$x=\cos y,$$
$$1=-y'\sin y,$$
$$0=-y''\sin y-y'^2\cos y,$$
$$\color{green}{y''=-\frac{\cos y}{\sin^3y}}.$$
