Finding $\sin\left(\cot^{-1}\frac{x}{2}\right)$, trouble understanding the solution Finding $\sin\left(\cot^{-1}\frac{x}{2}\right)$


I do not understand why they are considering only positive values. What has it to do with the range of $\cot^{-1}$ ? Could somebody dumb it down for me please Thanks!
 A: The range for the inverse cotangent is commonly taken to be $(0,\pi)$, where the sine is positive.
In general, we have
$$
\sin^2t=\frac{\sin^2t}{\sin^2t+\cos^2t}=\frac{1}{1+\cot^2t}
$$
dividing numerator and denominator by $\sin^2t$.
If $t=\cot^{-1}(x/2)$, you can then use
$$
\sin t=\frac{1}{\sqrt{1+\cot^2t}}
$$
Therefore
$$
\sin\cot^{-1}(x/2)=\frac{1}{\sqrt{1+\cot^2t}}=
\frac{1}{\sqrt{1+(x/2)^2}}
$$

If the range of the inverse cotangent is taken to be $(-\pi/2,0)\cup(0,\pi/2]$ the statement is surely wrong, because
$$
\frac{2}{\sqrt{4+x^2}}>0
$$
but $\sin\cot^{-1}(x/2)$ would be negative for $x<0$.
I've never seen that set being considered as the range of the inverse cotangent. The correct result in that case would be
$$
\sin\cot^{-1}\frac{x}{2}=
\begin{cases}
\dfrac{2}{\sqrt{4+x^2}} & \text{if $x\ge0$}\\[6px]
-\dfrac{2}{\sqrt{4+x^2}} & \text{if $x<0$}
\end{cases}
$$
But I can't think of a good reason for the inverse cotangent being 
defined that way, making it a non continuous function at $0$.
