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?

$\cos\left(\frac{3\pi}{2}+a\right)=\cos\left(\frac{3\pi}{2}\right)\cos(a)-\sin\left(\frac{3\pi}{2}\right)\sin(a)=-\sin(a)=-\frac{1}{3}=>\sin(a)=\frac{1}{3}$
$\cos(a)=\sqrt{1-\sin^2{a}}=\sqrt{1-\frac19}=\sqrt{\frac89}$
$\tan(a)=\frac{\frac{1}{3}}{\sqrt{\frac89}}=\frac{1}{\sqrt{8}}=\boxed{\frac{\sqrt{2}}{4}}$
