Question

How do you show that `cosx/(1 - tanx) - sin^2x/(cosx- sinx) = cosx + sinx`?

Answer 1

The following identities are important for this problem:

-`tantheta = sintheta/costheta`

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

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

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

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

Factor the numerator on the left as a difference of squares, `a^2 - b^2 = (a + b)(a - b)`.

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

`cosx + sinx = cosx + sinx`

Identity Proved!!

Hopefully this helps!

Answer 2

`LHS=cosx/(1-tanx)-sin^2x/(cosx-sinx)`

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

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

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

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

`=cosx+sinx=RHS`

Proved