## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

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 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}