Prove that: $\mathrm{cosec} \frac {\alpha}{8}...$ Prove that:
$$\mathrm{cosec}\frac {\alpha}{8}+\mathrm{cosec}\frac {\alpha}{4}+\mathrm{cosec}\frac {\alpha}{2}=\cot \frac {\alpha}{16} - \cot \frac {\alpha}{2}$$.
My Attempt:
$$\text{L.H.S}=\mathrm{cosec}\frac {\alpha}{8}+\mathrm{cosec}\frac {\alpha}{4} + \mathrm{cosec}\frac {\alpha}{2}$$
$$=\frac {1}{\sin(\alpha/8)} + \frac {1}{2\sin(\alpha/8)\cdot \cos(\alpha/8)} + \frac {1}{\sin (\alpha/2)}$$.
$$=\frac {2\cos (\alpha/8) +1}{2\sin(\alpha/8)\cdot\cos(\alpha/8)} + \frac {1}{\sin(\alpha/2)}$$.
How should I move on further? Please help.

Note that in general
$$\csc x + \cot x = \frac{1+\cos x}{\sin x} = \cot 2x.$$
Now, moving $\cot\frac{\alpha}{2}$ to the LHS and applying this identity multiple times gives
\begin{align}
  \csc\frac {\alpha}{8} + \csc\frac {\alpha}{4} + 
     \csc\frac {\alpha}{2} +  \cot \frac {\alpha}{2}
  &= \csc \frac{\alpha}{8} + \csc\frac{\alpha}{4} + \cot\frac{\alpha}{4} \\
  &= \csc \frac{\alpha}{8} + \cot\frac{\alpha}{8} \\
  &= \cot \frac{\alpha}{16}.
\end{align}
