## Precalculus (6th Edition) Blitzer

To verify the given identity, ${{\sin }^{2}}\theta \left( 1+{{\cot }^{2}}\theta \right)=1$ Recall Trigonometric Identities, \begin{align} & 1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta \\ & \csc \theta =\frac{1}{\sin \theta } \\ & \cot \theta =\frac{\cos \theta }{\sin \theta } \\ \end{align} Use the above identities and solve the left side of the given expression, \begin{align} & {{\sin }^{2}}\theta \left( 1+{{\cot }^{2}}\theta \right)={{\sin }^{2}}\theta \left( {{\csc }^{2}}\theta \right) \\ & ={{\sin }^{2}}\theta \cdot \frac{1}{{{\sin }^{2}}\theta } \\ & =1 \end{align} Therefore, ${{\sin }^{2}}\theta \left( 1+{{\cot }^{2}}\theta \right)=1$ Hence, it is proved that the given identity holds true.