<s>[INST] Knowing that $\cos(\frac{3\pi}{2}+\alpha)=-\frac{1}{3}$ and $\tan \alpha > 0$, find the exact value of $\tan \alpha$ First I tried to solve for $\alpha$:
$$\cos(\frac{3\pi}{2}+\alpha)=-\frac{1}{3} \Leftrightarrow \alpha = \arccos(-\frac{1}{3})+2k\pi-\frac{3\pi}{2} \lor \alpha = -\arccos(-\frac{1}{3})+2k\pi-\frac{3\pi}{2}$$
Then I put $\tan(\arccos(-\frac{1}{3})-\frac{3\pi}{2})$ and $\tan(-\arccos(-\frac{1}{3})-\frac{3\pi}{2})$ in the calculator and $\tan(\arccos(-\frac{1}{3})-\frac{3\pi}{2})$ was positive while the other was negative, so I assumed the positive one was the correct expression.
But I don't know how to simplify it to get the exact value. My book says the solution is $\frac{\sqrt{2}}{4}$. How do I solve this?
 [/INST] Hint: your first equation gives $\sin (\alpha)\,=\,\frac{1}{3}$ which can be easily solved to get $\tan(\alpha)\,=\,\frac{\sqrt2}{4}$
$$\cos(\frac{3\pi}{2}+\alpha)\,=\,-\sin(\alpha)$$
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