Question

Question #5cf65

Answer 1

Currently, as is, the identity is false. If we change `sinx - cosx` in the denominator on the left to `sinx + cosx`, it will be true though (thanks to Mrs. R. for pointing this out).

We know that `tantheta = sintheta/costheta`, `cottheta = 1/tantheta = 1/(sintheta/costheta) = costheta/sintheta`, `sectheta = 1/costheta` and `csctheta = 1/sintheta`. So:

`(cosx/sinx - sinx/cosx)/(sinx+ cosx) = 1/sinx - 1/cosx`

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

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

`(cosx - sinx)/(sinxcosx) = (cosx - sinx)/(cosxsinx)`

`LHS = RHS`

Identity proved!

Hopefully this helps!