Trigonometry (11th Edition) Clone

$$\frac{1}{\tan^2\alpha}+\cot\alpha\tan\alpha=\csc^2\alpha$$
$$A=\frac{1}{\tan^2\alpha}+\cot\alpha\tan\alpha$$ $$A=\Big(\frac{1}{\tan\alpha}\Big)^2+\cot\alpha\tan\alpha$$ 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 $A$: $$A=\cot^2\alpha+1$$ $$A=\csc^2\alpha\hspace{1cm}\text{(Pythagorean Identity)}$$