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?
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 : $$\frac{-\sqrt{3}}{3}$$
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 :
$$- \frac{1}{\sqrt{3}}$$