Prove that $\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}=\cot\left(\frac{3\pi}{2}-\alpha\right)$ Prove that $$\dfrac{\sin4\alpha}{1+\cos4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}=\cot\left(\dfrac{3\pi}{2}-\alpha\right)$$
The RHS is equal to $\tan\alpha,$ so we are to show $$\dfrac{\sin4\alpha}{1+\cos4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}=\tan\alpha$$
My try for simplifying the LHS: $$\dfrac{2\sin2\alpha\cos2\alpha}{1+\cos^22\alpha-\sin^22\alpha}\cdot\dfrac{\cos^2\alpha-\sin^2\alpha}{1+\cos^2\alpha-\sin^2\alpha}=\\=\dfrac{4\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)}{1+(\cos^2\alpha-\sin^2\alpha)^2-(2\sin\alpha\cos\alpha)^2}\cdot\dfrac{\cos^2\alpha-\sin^2\alpha}{1+\cos^2\alpha-\sin^2\alpha}\\=\dfrac{4\sin\alpha\cos\alpha(\cos\alpha-\sin\alpha)(\cos\alpha+\sin\alpha)}{1+\sin^4\alpha-2\cos^2\alpha\sin^2\alpha+\cos^4\alpha-4\sin^2\alpha\cos^2\alpha}\cdot\dfrac{\cos^2\alpha-\sin^2\alpha}{2\cos^2\alpha}$$
 A: Recall that $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$, and $\cos(x) = \frac{e^{ix}+e^{-ix}}{2},$ where $i = \sqrt{-1}$. Then:
$$X = \frac{\sin(4\alpha)}{1+\cos(4\alpha)}\cdot\frac{\cos(2\alpha)}{1+\cos(2\alpha)}=
\frac{1}{i}\cdot\frac{(e^{i4\alpha}-e^{-i4\alpha})(e^{i2\alpha}+e^{-i2\alpha})}{(2 + e^{i4\alpha}+e^{-i4\alpha})(2+e^{i2\alpha}+e^{-i2\alpha})}.$$
Let $s = e^{i2\alpha}$. Then:
$$X=\frac{1}{i}\cdot\frac{(s^2-s^{-2})(s + s^{-1})}{(2 + s^2+s^{-2})(2+s+s^{-1})} = \ldots \text{some boring algebra}\ldots = \\=\frac{1}{i}\frac{s-1}{s+1} = \frac{1}{i}\frac{e^{i2\alpha}-1}{e^{i2\alpha} + 1} = \\
\frac{1}{i}\frac{e^{i\alpha}(e^{i\alpha}-e^{-i\alpha})}{e^{i\alpha}(e^{i\alpha} + e^{-i\alpha})} = \frac{1}{i}\frac{e^{i\alpha}-e^{-i\alpha}}{e^{i\alpha} + e^{-i\alpha}}\\
= \displaystyle\frac{\frac{e^{i\alpha}-e^{-i\alpha}}{2i}}{\frac{e^{i\alpha} + e^{-i\alpha}}{2}} = \frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha).\\
$$
A: In fact, using
$$ 1-\cos 2x=2\sin^2x, 1+\cos2x=2\cos^2x $$
one has
\begin{eqnarray}
&&\dfrac{\sin4\alpha}{1+\cos4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}\\
&=&\dfrac{\sin4\alpha(1-\cos4\alpha)}{1-\cos^24\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}\\
&=&\dfrac{1-\cos4\alpha}{\sin4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}\\
&=&\dfrac{1-\cos4\alpha}{2\sin2\alpha}\cdot\dfrac{1}{1+\cos2\alpha}\\
&=&\dfrac{2\sin^22\alpha}{2\sin2\alpha}\cdot\dfrac{1}{2\cos^2\alpha}\\
&=&\dfrac{\sin2\alpha}{2\cos^2\alpha}=\tan\alpha\\
&=&\cot\left(\dfrac{3\pi}{2}-\alpha\right).
\end{eqnarray}
