Answer
The right side of the given equation is equivalent to $\sec{u}-\tan{u}$ 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 given equation.
Multiply $\dfrac{1-\sin{u}}{1-\sin{u}}$ to obtain:
\begin{align}
\text{RHS } &=\dfrac{\cos{u}}{1+\sin{u}} \times \dfrac{1-\sin{u}}{1-\sin{u}} \\[3mm]
&= \dfrac{\cos{u}(1-\sin{u})}{1-\sin^2{u}} \\[3mm]
&= \dfrac{\cos{u}(1-\sin{u})}{\cos^2{u}} \\[3mm]
&= \dfrac{1-\sin{u}}{\cos{u}} \\[3mm]
&= \dfrac{1}{\cos{u}}-\dfrac{\sin{u}}{\cos{u}} \\[3mm]
& = \sec{u}-\tan{u} \\[3mm]
&= \text{ LHS}
\end{align}
