Question

Prove that `(cosx+secx)^2 -= sec^2x+2+ cos^2x`?

Answer 1

Firstly, let us consider the RHS which we can expand and simplify:

`RHS = (cosx+secx)^2`
`\ \ \ \ \ \ \ \ = (cosx+secx)(cosx+secx)`
`\ \ \ \ \ \ \ \ = cosxcosx+cosxsecx+secxcosx+secxsecx`
`\ \ \ \ \ \ \ \ = cos^2x+2cosxsecx+sec^2x`
`\ \ \ \ \ \ \ \ = cos^2x+2cosx*1/cosx+sec^2x`
`\ \ \ \ \ \ \ \ = cos^2x+2+sec^2x`

Now, let us examine the LHS:

`LHS = sec^2x+2sin^2x+3cos^2x`
`\ \ \ \ \ \ \ \ = sec^2x+2sin^2x+2cos^2x + cos^2x`
`\ \ \ \ \ \ \ \ = sec^2x+2(sin^2x+cos^2x) + cos^2x`
`\ \ \ \ \ \ \ \ = sec^2x+2+ cos^2x`

Hence `LHS -= RHS` verifying the identity QED