Question

What is the derivative of `y = (secx + tanx)(secx -tanx)`?

Answer 1

`y' = 0`

Explanation:

Start by rewriting in terms of sine and cosine. For this problem we use the identities `sectheta = 1/costheta` ad `tantheta = sintheta/costheta`.

`y = (1/cosx + sinx/cosx)(1/cosx- sinx/cosx)`

`y = ((1 + sinx)/cosx)((1 - sinx)/cosx)`

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

`y = cos^2x/cos^2x`

`y = 1`

The derivative of any constant is `0`.

`y' = 0`

Hopefully this helps!