Answer
See the full explanation below.
Work Step by Step
$\frac{\tan 2\theta +\cot 2\theta }{\sec 2\theta }$
Apply the quotient identity of trigonometry $\tan \theta =\frac{\sin \theta }{\cos \theta }$ , and $\cot \theta =\frac{\cos \theta }{\sin \theta }$ , and the reciprocal identity of trigonometry $\sec \theta =\frac{1}{\cos \theta }$. Then the above expression can be further simplified as:
$\frac{\tan 2\theta +\cot 2\theta }{\sec 2\theta }=\frac{\frac{\sin 2\theta }{\cos 2\theta }+\frac{\cos 2\theta }{\sin 2\theta }}{\frac{1}{\cos 2\theta }}$
Multiply $\frac{\sin 2\theta }{\cos 2\theta }$ by $\frac{\sin 2\theta }{\sin 2\theta }$ , and $\frac{\cos 2\theta }{\sin 2\theta }$ by $\frac{\cos 2\theta }{\cos 2\theta }$.
$\begin{align}
& \frac{\frac{\sin 2\theta }{\cos 2\theta }+\frac{\cos 2\theta }{\sin 2\theta }}{\frac{1}{\cos 2\theta }}=\frac{\frac{\sin 2\theta }{\cos 2\theta }.\frac{\sin 2\theta }{\sin 2\theta }+\frac{\cos 2\theta }{\sin 2\theta }.\frac{\cos 2\theta }{\cos 2\theta }}{\frac{1}{\cos 2\theta }} \\
& =\frac{\frac{{{\sin }^{2}}2\theta }{\cos 2\theta \sin 2\theta }+\frac{{{\cos }^{2}}2\theta }{\cos 2\theta \sin 2\theta }}{\frac{1}{\cos 2\theta }} \\
& =\frac{\frac{{{\sin }^{2}}2\theta +{{\cos }^{2}}2\theta }{\cos 2\theta \sin 2\theta }}{\frac{1}{\cos 2\theta }}
\end{align}$
By using the Pythagorean identity of trigonometry ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, now the above expression can be further simplified as:
$\frac{\frac{{{\sin }^{2}}2\theta +{{\cos }^{2}}2\theta }{\cos 2\theta \sin 2\theta }}{\frac{1}{\cos 2\theta }}=\frac{\frac{1}{\cos 2\theta \sin 2\theta }}{\frac{1}{\cos 2\theta }}$
Then rewrite the main fraction bar as $\div $:
$\frac{\frac{1}{\cos 2\theta \sin 2\theta }}{\frac{1}{\cos 2\theta }}=\frac{1}{\cos 2\theta \sin 2\theta }\div \frac{1}{\cos 2\theta }$
Invert the divisor and multiply:
$\begin{align}
& \frac{1}{\cos 2\theta \sin 2\theta }\div \frac{1}{\cos 2\theta }=\frac{1}{\cos 2\theta \sin 2\theta }.\frac{\cos 2\theta }{1} \\
& =\frac{1}{\sin 2\theta }
\end{align}$
by using the reciprocal identity of trigonometry $csc\theta =\frac{1}{sin\theta }$, now the above expression can be further simplified as:
$\frac{1}{\sin 2\theta }=\csc 2\theta $
Thus, the left side of the expression is equal to the right side, which is
$\frac{\tan 2\theta +\cot 2\theta }{\sec 2\theta }=\csc 2\theta $.
