Answer
$$\frac{1}{\tan^2\alpha}+\cot\alpha\tan\alpha=\csc^2\alpha$$
Work Step by Step
$$A=\frac{1}{\tan^2\alpha}+\cot\alpha\tan\alpha$$
$$A=\Big(\frac{1}{\tan\alpha}\Big)^2+\cot\alpha\tan\alpha$$
According to a Reciprocal Identity:
$$\cot\alpha=\frac{1}{\tan\alpha}$$
so, $$\Big(\frac{1}{\tan\alpha}\Big)^2=\cot^2\alpha\hspace{1cm}(1)$$
Also, we can deduce that $$\cot\alpha\tan\alpha=1\hspace{1cm}(2)$$
Combine $(1)$ and $(2)$ into $A$:
$$A=\cot^2\alpha+1$$
$$A=\csc^2\alpha\hspace{1cm}\text{(Pythagorean Identity)}$$
