prove that(sin8xcosx-sin6xcos3x)/(cos2x.cosx-sin4xsin3x)=tan 2x

Asked by nimisha varghese | 23rd Aug, 2010, 12:00: AM

Expert Answer:

L.H.S = (sin8xcosx-sin6xcos3x)/(cos2xcosx-sin4xsin3x)
               Multiply numerator and denominator by 2 
        =  (2sin8xcosx-2sin6xcos3x)/(2cos2xcosx-2sin4xsin3x)
             Apply product to sum formulae 
        =  [(sin9x + sin7x) - (sin 9x + sin3x)]/[(cos3x + cosx) -(cosx - cos 7x)]
        =  (sin7x - sin3x) / (cos3x + cos7x)
           Apply sum to product formulae
         = (2cos5xsin2x) / (2cos5xcos2x)
         = tan2x = R.H.S

Answered by  | 23rd Aug, 2010, 09:44: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.