Answer
See the full explanation below.
Work Step by Step
${{\sec }^{2}}x{{\csc }^{2}}x$
Apply the Pythagorean identity of trigonometry ${{\sec }^{2}}x=1+{{\tan }^{2}}x$ , Then, the above expression can be further simplified as:
$\begin{align}
& {{\sec }^{2}}x{{\csc }^{2}}x=\left( 1+{{\tan }^{2}}x \right){{\csc }^{2}}x \\
& ={{\csc }^{2}}x+{{\tan }^{2}}x.{{\csc }^{2}}x
\end{align}$
By using the quotient identity of trigonometry ${{\tan }^{2}}x=\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}$ , and reciprocal identity of trigonometry ${{\csc }^{2}}x=\frac{1}{{{\sin }^{2}}x}$ , the above expression can be further simplified as:
$\begin{align}
& {{\csc }^{2}}x+{{\tan }^{2}}x.{{\csc }^{2}}x={{\csc }^{2}}x+\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}.\frac{1}{si{{n}^{2}}x} \\
& ={{\csc }^{2}}x+\frac{1}{{{\cos }^{2}}x}
\end{align}$
By using the reciprocal identity of trigonometry ${{\sec }^{2}}x=\frac{1}{{{\cos }^{2}}x}$ , Now, the above expression can be further simplified as:
$\begin{align}
& {{\csc }^{2}}x+\frac{1}{{{\cos }^{2}}x}={{\csc }^{2}}x+{{\sec }^{2}}x \\
& ={{\sec }^{2}}x+{{\csc }^{2}}x
\end{align}$
Thus, the left side of the expression is equal to the right side, which is
${{\sec }^{2}}x{{\csc }^{2}}x={{\sec }^{2}}x+{{\csc }^{2}}x$.
