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?

For the second one:


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*$2)$ 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*}
