How to prove the following identities? 
Prove: 
\begin{align}
  \tan(A) + \cot(A) & = 2 \text{cosec}(2A)\\
  \tan(45^{\circ}+A^{\circ}) - \tan(45^{\circ}-A^{\circ}) & = 2 \tan(2A^{\circ})\\
  \text{cosec}(2A) + \cot(2A) & = \cot(A)
  \end{align}

I have got all the formulas that I need but I just couldn't solve these. Some help, please?
 A: Hint
1) $\tan a+\cot a=\frac{\sin a}{\cos a}+\frac{\cos a}{\sin a}$. Bring to a common denominator and use the identities you know. (remember that $\csc x=\frac{1}{\sin x}$).
2) Use the same hint from 1. Use the formulas for $\sin(a+b)$ and $\cos(a+b)$.
3) Same hint from 1.
A: EDITED. In general there are two different techniques we can use to prove a trigonometric identity $A=B$. One is to transform one side into the other:
    $$A=A_1=A_2=\dots =A_n=B.$$
    The other is to look at the identity $A=B$ as a whole and convert it into an equivalent one and repeat the process until one known identity is found:
    $$A=B\Leftrightarrow A'=B'\Leftrightarrow A''=B''\Leftrightarrow\dots\Leftrightarrow A^{*}=B^{*}.$$
    The following hints are intended for proving your $3^{rd}$ identity by this second technique. Use
    $$\frac{\cos 2A}{\sin 2A}=\frac{2\cos ^{2}A-1}{2\sin A\cos A}=\frac{%
    \cos A}{\sin A}-\frac{1}{2\sin A\cos A}$$
    to obtain
    $$\csc 2A+\cot 2A=\cot A\Leftrightarrow \frac{1}{2\cos A}+\cos A-\frac{1}{%
    2\cos A}=\cos A.$$
Added. Proof:
$$\csc 2A+\cot 2A=\cot A\tag{1}$$
$$\begin{eqnarray*}
&\Leftrightarrow &\frac{1}{\sin 2A}+\frac{\cos 2A}{\sin 2A}=\frac{\cos A}{%
\sin A} \\
&\Leftrightarrow &\frac{\sin A}{\sin 2A}+\frac{\cos 2A}{\sin 2A}\sin A=\cos A
\\
&\Leftrightarrow &\frac{1}{2\cos A}+\left( \frac{\cos A}{\sin A}-\frac{1}{%
2\sin A\cos A}\right) \sin A=\cos A \\
&\Leftrightarrow &\frac{1}{2\cos A}+\left( \cos A-\frac{1}{2\cos A}\right)
=\cos A \\
\end{eqnarray*}$$
$$\Leftrightarrow \cos A=\cos A\tag{2},$$
which is an identity. Thus $(1)$ is also an identity.

Your $1^{st}$ identity can be proved by the first technique:
$$\begin{eqnarray*}
\tan A+\cot A &=&\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{\sin
^{2}A+\cos ^{2}A}{\cos A\sin A} \\
&=&\frac{1}{\cos A\sin A}=\dfrac{1}{\dfrac{\sin (2A)}{2}}=\cdots 
\end{eqnarray*}$$
A: For the second one:


*

*$2)$ Call $\alpha = \frac{\pi}{4} +A$ and $\beta= \frac{\pi}{4}-A$. So you have $\alpha + \beta = \frac{\pi}{2} \Rightarrow \cot(\alpha+\beta) = 0$. From this you have  $$\frac{1}{\tan(\alpha+\beta)} = \frac{1 - \tan{\alpha}\tan{\beta}}{\tan{\alpha} + \tan{\beta}}=0 \Rightarrow  \tan{\alpha} \cdot \tan{\beta}=1$$ Now note that $\alpha - \beta= 2A$ and so we again have 
\begin{align*}
\tan(\alpha-\beta) &= \frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\cdot \tan(\beta)} \\ &= \frac{\tan(\alpha)-\tan(\beta)}{2} \\ \Rightarrow \ 2\tan(2A)&= \tan(\alpha)- \tan(\beta)
\end{align*}

*$3)$ You have : \begin{align*} \csc{2A} + \cot{2A} &= \frac{1}{\sin{2A}} + \frac{\cos{2A}}{\sin{2A}} \\ & = \frac{1+\cos{2A}}{\sin{2A}} \\ &= \frac{1+\cos^{2}{A} - \sin^{2}{A}}{2 \cdot \sin{A} \cdot \cos{A}} \\ &= \frac{2 \cdot\cos^{2}{A}}{2 \cdot \sin{A} \cdot \cos{A}} = \cot(A)\end{align*}
