Answer
See the explanation below.
Work Step by Step
$\left( {{\cot }^{2}}\theta +1 \right)\left( {{\sin }^{2}}\theta +1 \right)$
We multiply the above expression as:
$\left( {{\cot }^{2}}\theta +1 \right)\left( {{\sin }^{2}}\theta +1 \right)={{\cot }^{2}}\theta {{\sin }^{2}}\theta +{{\cot }^{2}}\theta +{{\sin }^{2}}\theta +1$
By using the quotient identity of trigonometry ${{\cot }^{2}}\theta =\frac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }$ , the above expression can be further simplified as:
$\begin{align}
& {{\cot }^{2}}\theta {{\sin }^{2}}\theta +{{\cot }^{2}}\theta +{{\sin }^{2}}\theta +1=\frac{{{\cos }^{2}}\theta }{si{{n}^{2}}\theta }.si{{n}^{2}}\theta +{{\cot }^{2}}\theta +{{\sin }^{2}}\theta +1 \\
& ={{\cos }^{2}}\theta +{{\cot }^{2}}\theta +{{\sin }^{2}}\theta +1 \\
& ={{\cos }^{2}}\theta +{{\sin }^{2}}\theta +{{\cot }^{2}}\theta +1
\end{align}$
Apply the Pythagorean identity of trigonometry ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$; now, the above expression can be further simplified as:
$\begin{align}
& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +{{\tan }^{2}}\theta +1=1+{{\cot }^{2}}\theta +1 \\
& ={{\cot }^{2}}\theta +2
\end{align}$
Thus, the left side of the expression is equal to the right side, which is
$\left( {{\cot }^{2}}\theta +1 \right)\left( {{\sin }^{2}}\theta +1 \right)={{\cot }^{2}}\theta +2$.
