Find the exact value of $\tan⁡(\cos^{-1} (-\sqrt{3}/2))$. The question: Find the exact value of $\tan⁡(\cos^{-1} (-\sqrt{3}/2))$.

The link is the image of the method I used. However it isn't the right answer, how come this method doesn't work?
 A: Use the famous identity $$\tan(\arccos(x)) = \frac{\sqrt{1-x^2}}{x}$$ I think you can take it from your here. You can check the answer after you have done it. The answer is : $$\frac{-\sqrt{3}}{3}$$
A: The problem with your work is with your evaluation of the length of the vertical leg:
$\sqrt{2^2-(-\sqrt3)^2}=\sqrt{4-3}=\sqrt1=1$
This leads to the correct answer of $\tan\theta=\frac1{-\sqrt3}=-\frac{\sqrt3}3 $.
Also keep in mind that your assumption that the angle is in the second quadrant is not warranted, since cosine is also negative in the third quadrant.  In that case, since tangent is positive in the third quadrant, that would lead to an answer of $+\frac{\sqrt3}3$.
A: Follow :
$$ \tan (\arccos(- \frac{\sqrt{3}}{2}))$$
Suppose $x = \arccos(- \frac{\sqrt{3}}{2})$ :
$$ \cos x = - \frac{\sqrt{3}}{2} = \frac {\text{Adjacent}}{\text{hypotenuse}}$$
Pythagorean theorem:
$$\text{opposite side} = \pm~\sqrt{4-3} = \pm 1$$
$$ \tan x = \frac {\text{Opposite}}{\text{Adjacent}} =  \pm \frac{1}{\sqrt{3}} $$
Considering $ 0 \le x \le \pi$ . Take only negative as $\tan$ is negative in second quadrant :
$$- \frac{1}{\sqrt{3}}$$
