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?
 A: we have $$\frac{\sin(A)}{1+\cos(A)}+\frac{1+\cos(A)}{\sin(A)}=\frac{\sin^2(A)+(1+\cos(A))^2}{\sin(A)(1+\cos(A))}=\frac{2+2\cos(A)}{\sin(A)(1+\cos(A))}=$$
can you finish?
A: $$\frac{\sin^2A+(1+\cos(A))^2}{\sin(A)(1+\cos(A))} = \frac{\sin^2A+1+2\cos(A) + \cos^2(A)}{\sin(A)(1+\cos(A))}$$
Now, usse $$\sin^2(A) + \cos^2(A) = 1.$$
Take it from here.
A: [take 2 common in the numerator then you will get 2(1+cos(A))]
[in denominator there is a mistake it's 1+cos(A)]
A: 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.
A: $$\frac{\sin{A}}{1+\cos{A}}+\frac{1+\cos{A}}{\sin{A}}=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2}}+\frac{2\cos^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}=$$
$$=\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}+\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}=\frac{\sin^2\frac{A}{2}+\cos^2\frac{A}{2}}{\sin\frac{A}{2}\cos\frac{A}{2}}=\frac{2}{2\sin\frac{A}{2}\cos\frac{A}{2}}=\frac{2}{\sin{A}}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\i}{\mathrm{i}} \newcommand{\text}[1]{\mathrm{#1}} \newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}$
$$\frac{\sin(a)}{1+ cos(a
)}+\frac{1+ \cos(a
)}{\sin(a)}=2 cosec(a)$$
$$\frac{\sin^2a + (1+\cos a)^2}{\sin a (1+ \cos a)} = \frac{2}{\sin a}$$
$$\frac{\sin^2a + (1+\cos^2 a + 2\cos a)}{\sin a (1+ \cos a)} = \frac{2}{\sin a}$$
$$\frac{2+2\cos a}{\sin a (1+ \cos a)} = \frac{2}{\sin a}$$
$$\bbx{\frac{1+1\cos a}{1 (1+ \cos a)} = \frac{1}{1}}$$
A: Alternatively:
$$\frac{\sin A(1-\cos A)}{\sin^2 A}+\frac{1+\cos A}{\sin A}=2cosec A.$$
