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:
$\begin{eqnarray*} \tan A+\cot A &=&\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{\sin ^{2}A+\cos ^{2}A}{\cos A\sin A} \\ &=&\frac{1}{\cos A\sin A}=\dfrac{1}{\dfrac{\sin (2A)}{2}}=\cdots \end{eqnarray*}$