Note that
$$\lim_{x\to -2^+}\frac{\sqrt{4-x^2}}{1-x}=0\;\;,\;\;\lim_{x\to 1^-}\frac{\sqrt{4-x^2}}{1-x}=\infty$$
and thus
$$\left\{\alpha\in\Bbb R\;;\;\alpha=\log\frac{\sqrt{4-x^2}}{1-x}\;,\;\;x\in(-2,1)\right\}=\Bbb R$$
and from here
$$\text{Im}\left(\sin\log\frac{\sqrt{4-x^2}}{1-x}\right)=[-1,1]$$
Using $\displaystyle |\sin x|=\left\{\begin{matrix}
\displaystyle \sin x \;,& \displaystyle 0 \leq x\leq \frac{\pi}{2} \\\\
\displaystyle \sin x \;,& \displaystyle \frac{\pi}{2} \leq x\leq \pi \\\\
\displaystyle -\sin x\;, & \displaystyle\pi \leq x\leq \frac{3\pi}{2} \\\\
\displaystyle -\sin x \;,& \displaystyle\frac{3\pi}{2} \leq x\leq 2\pi
\end{matrix}\right.$ and $\displaystyle |\cos x|=\left\{\begin{matrix}
\displaystyle \cos x \;,& \displaystyle 0 \leq x\leq \frac{\pi}{2} \\\\
\displaystyle -\cos x \;,& \displaystyle \frac{\pi}{2} \leq x\leq \pi \\\\
\displaystyle -\cos x \;,& \displaystyle\pi \leq x\leq \frac{3\pi}{2} \\\\
\displaystyle \cos x \;,& \displaystyle\frac{3\pi}{2} \leq x\leq 2\pi
\end{matrix}\right.$
So $$\displaystyle f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}} = \sin x\cdot \left|\cos x\right|-\cos x\cdot \left|\sin x\right|$$
Here Function $f(x)$ is Periodic With Time Period $= 2\pi.$
So we will Calculate for Only one Time Period.
In $\bullet \displaystyle \; 0 \leq x\leq \frac{\pi}{2}\;,$ We get $f(x) = \sin x\cdot \cos x-\cos x\cdot \sin x = 0$
In $\bullet \displaystyle \; \frac{\pi}{2} \leq x\leq \pi\;,$ We get $f(x)=-\sin 2x .$ So $0 \leq f(x)\leq 1$
In $\bullet \displaystyle \; \pi \leq x\leq \frac{3\pi}{2}\;,$ We get $f(x)=0 .$
In $\bullet \displaystyle \; \frac{3\pi}{2} \leq x\leq 2\pi\;,$ We get $f(x)=\sin 2x .$ So $-1 \leq f(x)\leq 0$
So Here We get $\displaystyle -1 \leq f(x)\leq 1$
Best Answer
Note that $\sqrt{\log(\cos(\sin(x)))}$ is defined if and only if $\cos(\sin(x))\geq 1$ otherwise we have the square root of a negative number. Now since the cosine function is upper bounded by $1$, $\cos(\sin(x))$ has to be $1$ (such equality holds for example when $x=0$). So what is the range of the given function?