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 is ${{\left( \tan \theta +\cot \theta \right)}^{2}}$ and it can be simplified by using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. After that, we use the quotient identity $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and Pythagorean identity $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $. Thus, ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$ and $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta $; therefore, ${{\cot }^{2}}\theta ={{\csc }^{2}}\theta -1$.
$\begin{align}
& {{\left( \tan \theta +\cot \theta \right)}^{2}}={{\tan }^{2}}\theta +2.\tan \theta .\cot \theta +{{\cot }^{2}}\theta \\
& ={{\sec }^{2}}\theta -1+2.\frac{\sin \theta }{\cos \theta }.\frac{\cos \theta }{\sin \theta }+{{\csc }^{2}}\theta -1 \\
& ={{\sec }^{2}}\theta -1+2+{{\csc }^{2}}\theta -1 \\
& ={{\sec }^{2}}\theta +{{\csc }^{2}}\theta
\end{align}$
Hence, the expression on the left side is equal to the expression on the right side.
