## Precalculus (6th Edition) Blitzer

We know that, $\sin \theta =\frac{1}{\csc \theta }$. Therefore, we can rewrite the equation: $\sin \theta \csc \theta -{{\cos }^{2}}\theta$, as: \begin{align} & \sin \theta \csc \theta -{{\cos }^{2}}\theta =\sin \theta \left( \frac{1}{\sin \theta } \right)-{{\cos }^{2}}\theta \\ & =1-{{\cos }^{2}}\theta \\ & ={{\sin }^{2}}\theta \end{align} Therefore, from (1) we get; $\sin \theta \csc \theta -{{\cos }^{2}}\theta ={{\sin }^{2}}\theta$