Typically, such simple relations can be derived from either direction using basic definitions. Let's prove the first one starting from right side and the second one from the left side. $$1+\tan^2\alpha=1+\frac{\sin^2\alpha}{\cos^2\alpha}=\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}=\frac{1}{\cos^2\alpha}=\sec^2\alpha$$ and $$\csc^2\alpha=\frac{1}{\sin^2\alpha}=\frac{\sin^2\alpha + \cos^2\alpha}{\sin^2\alpha}=1+\frac{\cos^2\alpha}{\sin^2\alpha}=1+\cot^2\alpha$$