## Precalculus (6th Edition) Blitzer

The expression on the left side ${{\sin }^{2}}\theta \left( 1+{{\cot }^{2}}\theta \right)$ can be simplified by multiplying the terms and then using the quotient identity ${{\cot }^{2}}x=\frac{{{\cos }^{2}}x}{{{\sin }^{2}}x}$ and Pythagorean identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ \begin{align} & {{\sin }^{2}}\theta \left( 1+{{\cot }^{2}}\theta \right)={{\sin }^{2}}\theta +{{\sin }^{2}}\theta .{{\cot }^{2}}\theta \\ & ={{\sin }^{2}}\theta +{{\sin }^{2}}\theta .\frac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta } \\ & ={{\sin }^{2}}\theta +{{\cos }^{2}}\theta \\ & =1 \end{align} Thus, the expression on the left side is equal to the expression on the right side ${{\sin }^{2}}\theta \left( 1+{{\cot }^{2}}\theta \right)=1$.