Question

How do you evaluate `sec^2(18) - tan^2(18)`?

Answer 1

`sec^2(18) - tan^2(18)`

`= 1/cos^2(18) - sin^2(18)/cos^2(18)`

`= (1-sin^2(18))/cos^2(18)`

`= cos^2(18)/cos^2(18)`

`= 1`