Answer
The expression on the left side is equal to the expression on the right side
Work Step by Step
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 ${{\sin }^{2}}\theta \left( 1+{{\cot }^{2}}\theta \right)=1$.
