Đáp án:
Giải thích các bước giải:
$ (tan3x + tanx)(tan3x - tanx)$
$ = (\dfrac{sin3x}{cos3x} + \dfrac{sinx}{cosx})(\dfrac{sin3x}{cos3x} - \dfrac{sinx}{cosx})$
$ = (\dfrac{sin3xcosx + cos3xsinx}{cosxcos3x}).(\dfrac{sin3xcosx - cos3xsinx}{cosxcos3x})$
$ = \dfrac{sin(3x + x)}{cosxcos3x}.\dfrac{sin(3x - x)}{cosxcos3x}$
$ = \dfrac{sin4xsin2x}{cos²xcos²3x} = \dfrac{2cos2xsin²2x}{cos²xcos²3x}$
$ = \dfrac{8cos2xsin²xcos²x}{cos²xcos²3x} = \dfrac{8cos2xcos²x}{cos²3x}.$