Answer
$$\sin\Big(\cos^{-1}(-\frac{1}{2})+\tan^{-1}(-\sqrt3)\Big)=-\frac{\sqrt3}{2}$$
Work Step by Step
$$\sin\Big(\cos^{-1}(-\frac{1}{2})+\tan^{-1}(-\sqrt3)\Big)$$
1) $\cos^{-1}(-\frac{1}{2})$
$-\frac{1}{2}\lt0$ means that you can find the answer in either quadrant II or III, the area of negativity for a cosine function.
In quadrant II, there is this value of $\frac{2\pi}{3}$ that $\cos\frac{2\pi}{3}=-\frac{1}{2}$.
So, $$\cos^{-1}(-\frac{1}{2})=\frac{2\pi}{3}$$
2) $\tan^{-1}(-\sqrt3)$
$-\sqrt3\lt0$ means that you can find the answer in either quadrant II or IV, the area of negativity for a tangent function.
In quadrant II, $\frac{2\pi}{3}$ is also the angle that $\tan\frac{2\pi}{3}=-\sqrt3$.
Therefore, $$\tan^{-1}(-\sqrt3)=\frac{2\pi}{3}$$
3) Refer back to the original formula
$$\sin\Big(\cos^{-1}(-\frac{1}{2})+\tan^{-1}(-\sqrt3)\Big)$$
$$=\sin\Big(\frac{2\pi}{3}+\frac{2\pi}{3}\Big)$$
$$=\sin\frac{4\pi}{3}$$
$\frac{4\pi}{3}$ lies in quadrant III, and corresponds to angle $\frac{\pi}{3}$ in quadrant I. However, quadrant III is where $\sin\theta\lt0$. Therefore,
$$=-\sin\frac{\pi}{3}$$
$$=-\frac{\sqrt3}{2}$$
