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