Question

How do you verify `sin x + cos x = (2sin^2 x - 1) /(sin x -cos x)`?

Answer 1

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

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

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

Here it looks like we're at a dead end, but in fact we're not.

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

`sin^2x - cos^2x = 1 - 2cos^2x`

`1 - cos^2x - cos^2x = 1 - 2cos^2x`

`1 - 2cos^2x = 1 - 2cos^2x -> "Identity proved!"`

Hopefully this helps!

Answer 2

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

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

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

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

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

`=(sinx+cosx)=LHS`