Proving trigonometric identity $\frac{\sin(A)}{1+ \cos(A )}+\frac{1+ \cos(A )}{\sin(A)}=2 \csc(A)$ 
$$
\frac{\sin(A)}{1+\cos(A)}+\frac{1+\cos(A)}{\sin(A)}=2\csc(A)
$$

\begin{align}
\mathrm{L.H.S}&= \frac{\sin^2A+(1+\cos^2(A))}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{\sin^2A+2\sin(A)\cos(A)+\cos^2(A)+1}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{2+2\sin(A)\cos(A)}{\sin(A)(1+\cos(A))}
\end{align}
What should be done from here?

You make several mistakes, the main one being
$$
(a+b)^2=a^2+b^2
$$
The mistake is $(1+\cos(A))^2=1+\cos^2(A)$, whereas it should be
$$
(1+\cos(A))^2=1+2\cos(A)+\cos^2(A)
$$
Note that
$$
\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}
$$
where $a=\sin(A)$ and $b=1+\cos(A)$.
In the second step you also arbitrarily insert a term $2\sin(A)\cos(A)$ with no justification.
Start again.
