Question

How do you prove `(sec+tan)(sec-tan)=1`?

Answer 1

We will be using the following:

• `(a+b)(a-b) = a^2-b^2`

• `tan^2(theta)+1 = sec^2(theta)`

With those:

`(sec(theta)+tan(theta))(sec(theta)-tan(theta))`

`=sec^2(theta)-tan^2(theta)`

`=(tan^2(theta)+1)-tan^2(theta)`

`=1`