Proving a trigonometric relation $$\frac{1+\cos{2\alpha}}{\sin^2{2\alpha}}=\frac{1}{2}\csc^2{\alpha}$$
Here the question is that I can prove that the $LHS=RHS$ if I use the variable $x$ but if we take
($2x=\pi$ on the left side) and ($x=\dfrac{\pi}{2}$ on the other side)
then the equation is not equal,why?
Then prove:
\begin{align}
\frac{1+\cos{2\alpha}}{\sin^2{2\alpha}} & = \frac{1+\cos{2\alpha}}{1-\cos^2{2\alpha}} \\
\require{cancel}& =\frac{\cancel{(1+\cos{2\alpha})}}{\cancel{(1+\cos{2\alpha})}(1-\cos{2\alpha})} \\ 
& =\frac{1}{1-\cos{2\alpha}}\\
& =\frac{1}{2\sin^2{\alpha}} & \text{because}, 1-\cos{2\alpha}=2\sin^2{\alpha}\\
& =\frac{1}{2}\csc^2{\alpha}\\
\end{align}
Actually it was an indefinite integration but I just noticed this.

$$
\frac {1+\cos 2x}{\sin^2 2x} = \frac {2\cos^2 x}{4 \sin^2 x \cos^2 x} = \frac 1{2\sin^2 x} = \frac 12 \csc^2 x
$$
