Prove $\frac{1}{\sin{2\alpha}}+\frac{1}{\sin{2^2\alpha}}+\cdots+\frac{1}{\sin{2^n\alpha}}=\frac{1}{\tan{\alpha}}-\frac{1}{\tan{2^n\alpha}}$ Can someone help me with the problem?
Let $\alpha \in \mathbb{R} \backslash \{2^{-m}k\pi; k\in\mathbb{Z},m\in\mathbb{N} \}$. 
Prove that the following equation is true for  $\forall n\in N$:
$$\frac{1}{\sin{2\alpha}}+\frac{1}{\sin{2^2\alpha}}+\cdots+\frac{1}{\sin{2^n\alpha}}=\frac{1}{\tan{\alpha}}-\frac{1}{\tan{2^n\alpha}}$$

This can be shown by Induction ( I will leave you to show the base case).
Observe that
\begin{eqnarray*}
\frac{1}{\sin (2^{n+1} \alpha)} &= & \frac{1}{ 2 \sin (2^{n} \alpha) \cos (2^{n} \alpha)} \\ 
&= & \frac{1}{ 2 \cos^2 (2^{n} \alpha) \tan (2^{n} \alpha)} \\ 
\end{eqnarray*}
Now add this term on 
\begin{eqnarray*}
-\frac{1}{\tan (2^{n} \alpha)} +\frac{1}{\sin (2^{n+1} \alpha)} &= & -\frac{1}{\tan (2^{n} \alpha)} \left(  \frac{1}{ 2 \cos^2 (2^{n} \alpha) } -1\right)  \\ 
&= & \frac{1 -2 \cos^2 (2^{n} \alpha) }{ 2 \cos^2 (2^{n} \alpha) \sin (2^{n} \alpha)} \\ 
&= &  -\frac{1 }{  \tan (2^{n+1} \alpha)}. \\ 
\end{eqnarray*}
