Recall angle addition formulae:
#sin(A+B)=sinAcosB+cosAsinB#
#sin(A-B)=sinAcosB-cosAsinB#
#tan(A+B)=(tanA+tanB)/(1-tanAtanB)#
#tan(A-B)=(tanA-tanB)/(1+tanAtanB)#
Recall also the Pythagorean identity:
#sin^2x+cos^2x=1#
which tells us
#|sinA|=15/17#
#|cosB|=3/5#
The given angular interval for both #A# and #B#, #((3pi)/2,2pi)#, corresponds to the fourth quadrant of the Cartesian plane. We use this to deduce the signs of #sinA# and #cosB# - sine is negative in this quadrant and cosine positive. So
#sinA=-15/17#
#cosB=3/5#
and thus
#tanA=-15/8#
#tanB=-4/3#
We now have all the knowledge we need to answer the specific questions:
a)
#sin(A+B)=sinAcosB+cosAsinB=#
#-15/17*3/5+8/17*(-4/5)=-77/85#
b)
#sin(A-B)=sinAcosB-cosAsinB=#
#-15/17*3/5-8/17*(-4/5)=-13/85#
c)
#tan(A+B)=(tanA+tanB)/(1-tanAtanB)=#
#(-15/8-4/3)/(1-(-15/8)*(-4/3))=(-77/24)/(-36/24)=77/36#
d)
#tan(A-B)=(tanA-tanB)/(1+tanAtanB)=#
#(-15/8-(-4/3))/(1+(-15/8)*(-4/3))=(-13/24)/(84/24)=-13/84#
e)
#sin(A+B)# is negative, so we are either in quadrant 3 or 4.
#tan(A+B)# is positive, so we are either in quadrant 1 or 3.
Therefore #A+B# is in quadrant 3.
f)
#sin(A-B)# is negative, so we are either in quadrant 3 or 4.
#tan(A-B)# is negative, so we are either in quadrant 2 or 4.
Therefore #A-B# is in quadrant 4.
