<s>[INST] Trigonometric Identities: Given $\tan(2a)=2$ and $\frac{3\pi}{2}Given $\tan(2a)=2$ and $\frac{3\pi}{2}<a<2\pi$ find value of $\tan(a)$
I first found the values of $\cos(2a)$ and $\sin(2a)$ and then used the half angle formula.
$$\tan(a)=\tan\frac{2a}{2}=\frac{1-\cos(2a)}{\sin(2a)}
\implies\frac{5}{2\sqrt 5}\left(1-\frac{\sqrt{5}}5\right)$$
I then simplified that to $\frac{\sqrt{5}}{2}-1$
Then because it is in fourth quadrant I multiplied by negative 1 and got:
$$1-\frac{\sqrt{5}}{2}$$
Am I doing something wrong, or am I correct?
 [/INST] You made two mistakes, but your method does work.
If $\tan(2a)=2$, then either
Case 1:  $0< 2a +2 n \pi<\pi/2$, $\cos(2a)=1/\sqrt{5}$, and $\sin(2a)= 2/\sqrt{5}$ (first quadrant),  or
Case 2:  $\pi<2a+2 n \pi<3 \pi/2$, $\cos(2a)=-1/\sqrt{5}$, and $\sin(2a)= -2/\sqrt{5}$ (third quadrant)
for some integer $n$.  
The problem stated that $3 \pi/2 < a < 2 \pi$, so  $3 \pi < 2 a < 4 \pi$ and that puts us in Case 2 with $n=-1$.  
Now using your method
$$\tan(a)=\frac{1 - \cos(2 a)}{\sin(2 a)} = \frac{1+1/\sqrt{5}}{-2/\sqrt{5}}
=  - \frac{\sqrt{5}+1}{2}.
$$
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