Question

How do you show that `sec(2x) + tan(2x) + 1 = 2/(1 - tanx)`?

Answer 1

First of all, `sectheta = 1/costheta` and `tantheta = sintheta/costheta`.

`1/(cos2x) +( sin2x)/(cos2x) + 1 = 2/(1 - sinx/cosx)`

`(sin2x + 1)/(cos2x) + 1 = 2/((cosx - sinx)/cosx)`

`(sin2x + 1)/(cos2x) + (cos2x)/(cos2x) = 2/((cosx - sinx)/cosx)`

We now establish the double angle identities. I won't go into the proofs for these, but I would just like to emphasize that these formulas are very important.

•`sin2theta = 2sinthetacostheta`
•`cos2theta = 1 - 2sin^2theta = 2cos^2theta - 1 = cos^2theta - sin^2theta`

`(2sinxcosx + 1 + cos2x)/(cos2x) = (2cosx)/(cosx - sinx)`

`(2sinxcosx + 2cos^2x - 1 + 1)/(cos^2x - sin^2x) = (2cosx)/(cosx- sinx)`

Factor:

`(2cosx(sinx + cosx))/((cosx + sinx)(cosx - sinx)) = (2cosx)/(cosx - sinx)`

`(2cosx)/(cosx + sinx) = (2cosx)/(cosx - sinx)`

Identity proved!

Hopefully this helps!