Answer
The righ side of the equation is eqiuvalent to $\csc{u} - \cot{u}$ therefore the given equation is an identity.
Refer to the solution below,
Work Step by Step
Work on the right hand side (RHS) of the equation.
Multiply the RHS by $\dfrac{1-\cos{u}}{1-\cos{u}}$ to obtain:
\begin{align}
\text{RHS } &= \dfrac{\sin{u}}{1+\cos{u}} \times \dfrac{1-\cos{u}}{1-\cos{u}} \\[3mm]
&= \dfrac{\sin{u}(1-\cos{u})}{1-\cos^2{u}} \\[3mm]
&= \dfrac{\sin{u}(1-\cos{u})}{\sin^2{u}} \\[3mm]
&= \dfrac{1-\cos{u}}{\sin{u}} \\[3mm]
&= \dfrac{1}{\sin{u}}-\dfrac{\cos{u}}{\sin{u}}\\[3mm]
&= \csc{u}-\cot{u} \\[3mm]
&= \text{ LHS}
\end{align}
