<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] Just use the double-angle identity.
$$\tan(2a) = \frac{2\tan{a}}{1-\tan^2(a)}$$
$$\implies \frac{2\tan{a}}{1-\tan^2(a)} = 2$$
$$\implies 2\tan{a} = 2-2\tan^2{a} \implies \tan{a} = 1-\tan^2{a} \implies \tan^2{a}+\tan{a}-1 = 0$$
Set $t = \tan{a}$ and solve for $t$.
$$t^2+t-1 = 0$$
$$t = \frac{-1\pm\sqrt{5}}{2}$$
Plug in $\tan{a}$.
$$\tan{a} = \frac{-1\pm\sqrt{5}}{2}$$
But $\frac{3\pi}{2} < a < 2\pi$, so $\tan{a} < 0$.
$$\tan{a} = \frac{-1-\sqrt{5}}{2}$$
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