Show that [(cosx-sinx)^2]/cosx=secx-2sinx

[(cosx-sinx)^2]/cosx=secx-2sinx

= [cos^2x-2cosxsinx+sin^2x]/cosx

= [1-2cosxsinx]/cosx

= 1/cosx- 2cosxsinx/cosx

=sexc-2sinx

Is it acceptable? Thank You.

YES

Well done, you just forgot to cancel the cos x in "-2cosxsinx/cosx"