we have
cos3x/(sinxcosx)
we know that
cos(3x) can be written as
cos(2x +x)=[cos2x*sinx-sin2x*cosx]
Using this value of cos(3x) in given equation
[cos2x*sinx-sin2x*cosx]/(sinxcosx)
[cos2x*sinx]/(sinxcosx)-[sin2x*cosx]/(sinxcosx)
[cos2x]/(cosx)-[sin2x]/(sinx)---------> secx*cos2x-csecx*sin2x
the answer is the option B.) secxcos2x-cscxsin2x