Answer: see proof below
Explanation:
[tex]sec^2\alpha=\dfrac{1}{1-\sin^2 \alpha}[/tex]
LHS → RHS
[tex]sec^2\alpha\qquad \text{Given}\\\\\dfrac{1}{cos^2\alpha}\qquad \text{sec is reciprocal of cos}\\\\\\\dfrac{1}{1-sin^2\alpha}\qquad \text{Identity}:cos^2\alpha + sin^2\alpha = 1\\\\\\\dfrac{1}{1-sin^2\alpha}=\dfrac{1}{1-sin^2\alpha}\quad \checkmark[/tex]