Question

Question #0b5d8

Answer 1

Proof below

Explanation:

First know these three identities:
`sin2x=2sinxcosx`
`cos2x=cos^2x-sin^2x`
`sin^2x+cos^2x=1`

First, we change the `sin2x` and `cos2x`
`(1-sin2x)/(cos2x)=(1-2sinxcosx)/(cos^2x-sin^2x)`

Now this part may be hard to see, but you need to split the `1` in the numerator using the third identity
`=(cos^2x+sin^2x-2sinxcosx)/(cos^2x-sin^2x)`

Rearrange to look more like the expansion of a square
`=(cos^2x-2sinxcosx+sin^2x)/(cos^2x-sin^2x)`

Factorise the numerator, while using the difference two swaures the factor the denominator
`=(cosx-sinx)^2/((cosx-sinx)(cosx+sinx))`

Cancel like terms:
`=(cosx-sinx)/(cosx+sinx)`

And we're done