Question

How do you prove `(secx-tanx)(secx+tanx) =secx`?

Answer 1

The given identity is false.

`(sec(x) - tan(x))(sec(x) + tan(x)) = 1`

Explanation:

We will be using the following:

• `sec(x) = 1/cos(x)` (by definition)

• `tan(x) = sin(x)/cos(x)` (by definition)

• `(a-b)(a+b) = a^2 - b^2` (difference of squares formula)

• `sin^2(x) + cos^2(x) = 1` (identity)

---

`(sec(x) - tan(x))(sec(x) + tan(x)) = sec^2(x) - tan^2(x)`
(by the difference of squares formula)

`= (1/cos(x))^2 - (sin(x)/cos(x))^2`
(by definition of secant and tangent)

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

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

`= cos^2(x)/cos^2(x)`
(by the above identity)

`=1`