Answer
$$\tan(-105^\circ)=2+\sqrt 3$$
D is the correct answer.
Work Step by Step
$$\tan(-105^\circ)$$
We see that $$\tan(-105^\circ)=\frac{\sin(-105^\circ)}{\cos(-105^\circ)}$$
and $\sin(-105^\circ)=-\sin105^\circ$ and $\cos(-105^\circ)=\cos105^\circ$. That means, $$\tan(-105^\circ)=\frac{-\sin(105^\circ)}{\cos(105^\circ)}=-\tan105^\circ$$
So now we only need to find the value of $\tan105^\circ$, by taking the sum of $60^\circ$ and $45^\circ$. $$\tan105^\circ=\tan(60^\circ+45^\circ)$$ $$=\frac{\tan60^\circ+\tan45^\circ}{1-\tan60^\circ\tan45^\circ}$$ $$=\frac{\sqrt 3+1}{1-1\times\sqrt 3}$$ $$=\frac{\sqrt 3+1}{1-\sqrt 3}$$ $$=\frac{\sqrt 3+1}{1-\sqrt 3}\times\frac{1+\sqrt 3}{1+\sqrt 3}$$ $$=\frac{(\sqrt 3+1)^2}{1-3}$$ $$=\frac{4+2\sqrt 3}{-2}$$ $$=-2-\sqrt 3$$
Therefore, $\tan(-105^\circ)=-\tan105^\circ=2+\sqrt 3$
D is the correct answer.
