Question

Question #4a238

Answer 1

`LHS=(sin3x-2cos2x-sinx)/(cos3x+cosx)`

`=(sin3x-sinx-2cos2x)/(cos3x+cosx)`

`=(2cos2xsinx-2cos2x)/(2cos2xcosx)`

`=(2cos2x(sinx-1))/(2cos2xcosx)`

`=sinx/cosx-1/cosx`

`=tanx-secx=RHS`