Answer
The left side of the equation is equivalent to $\dfrac{\cot{v}+1}{\cot{v}-1}$ therefore the given equation is an identity.
Refer to the solution below.
Work Step by Step
Work on the left hand side (LHS).
Divide by } the nnumerator and denomoinator of the LHS by $\tan{v}$ to obtain:
\begin{align}
\text{LHS } & = \dfrac{\dfrac{1}{\tan{v}}+1}{\dfrac{1}{\tan{v}}-1} \\[3mm]
&= \dfrac{\cot{v}+1}{\cot{v}-1} \\[3mm]
&= \text{ RHS}
\end{align}
