How do you prove $\tan p + \cot p = 2\csc 2p$?

Hint: Given equation is a trigonometric identity which means that the equality will hold true for any value of the angle $p$ . Proving the given identity means simplifying or transforming the expression in LHS (or RHS) in such a way that it resembles the expression in the RHS (or LHS). Here we will have to also use the half-angle formula to relate double-angle $2p$ to angle $p$.

Formula used:
$\tan p = \dfrac{{\sin p}}{{\cos p}}$
$\cot p = \dfrac{1}{{\tan p}} = \dfrac{{\cos p}}{{\sin p}}$
${\sin ^2}p + {\cos ^2}p = 1$
$\sin 2p = 2\sin p\cos p$ [half-angle formula or identity]

Complete step-by-step solution:
We have to prove that the expression $\tan p + \cot p$ is equal to the expression $2\csc 2p$.
In the LHS we have $\tan p + \cot p$.
We know that $\tan p = \dfrac{{\sin p}}{{\cos p}}$. And also $\cot p = \dfrac{1}{{\tan p}} = \dfrac{{\cos p}}{{\sin p}}$.
So we can write the LHS as,
$\tan p + \cot p = \dfrac{{\sin p}}{{\cos p}} + \dfrac{{\cos p}}{{\sin p}}$
Now we can simplify the expression as simple addition of fraction,
$\dfrac{{\sin p}}{{\cos p}} + \dfrac{{\cos p}}{{\sin p}} = \dfrac{{(\sin p \times \sin p) + (\cos p \times \cos p)}}{{\cos p \times \sin p}} = \dfrac{{{{\sin }^2}p + {{\cos }^2}p}}{{\sin p\cos p}}$
Since ${\sin ^2}p + {\cos ^2}p = 1$, we have,
$\dfrac{{{{\sin }^2}p + {{\cos }^2}p}}{{\sin p\cos p}} = \dfrac{1}{{\sin p\cos p}}$
Now we multiply both the numerator and denominator by $2$,
$\dfrac{1}{{\sin p\cos p}} \times \dfrac{2}{2} = \dfrac{2}{{2\sin p\cos p}}$
Using half-angle formula $\sin 2p = 2\sin p\cos p$, we can write,
$\dfrac{2}{{2\sin p\cos p}} = \dfrac{2}{{\sin 2p}}$
Also we know that $\dfrac{1}{{\sin p}} = \csc p$. So we can write,
$\dfrac{2}{{\sin 2p}} = 2\csc 2p$
Thus, we get the LHS in the form $2\csc 2p$.
Also the RHS given in the question is $2\csc 2p$.
Therefore, LHS = RHS.

Hence, we proved that $\tan p + \cot p = 2\csc 2p$.

Note: The above discussed solution may not be the only way to prove the given identity. We can prove an identity by various ways using different trigonometric properties or identities. Also, we can choose to transform the RHS to make it resemble the LHS. In case we get confused as to which identity to use where, we can transform both LHS and RHS to simpler terms using basic trigonometric properties or identities like we used to convert $\tan p$ and $\cot p$ in terms of $\sin p$ and $\cos p$ in this solution.