As #tanA=-4/3# and #pi/2 < A < pi#, #/_A# lies in Quadrant 2
As #cosB=5/6# and #0 < B < pi/2#, #/_B# lies in Quadrant 2.
Hence #secB=1/(5/6)=6/5#
and #tanB=sqrt(sec^2B-1)=sqrt((6/5)^2-1)=sqrt(11/25)=sqrt11/5# and it is positive as #/_B# lies in Quadrant 2.
#:.tan(A-B)=(tanA-tanB)/(1+tanAtanB)#
= #(-4/3-sqrt11/5)/(1-4/3xxsqrt11/5)#
= #(-20-3sqrt11)/(15-4sqrt11)#
(multiplying numerator and denominator by #15#)
Let us now rationalize the denominator by multiplying numerator and denominator by #15+4sqrt11#, and we get
#tan(A-B)=(-20-3sqrt11)/(15-4sqrt11)xx(15+4sqrt11)/(15+4sqrt11)#
= #(-300-80sqrt11-45sqrt11-12xx11)/(15^2-(4sqrt11)^2)#
= #(-300-125sqrt11-132)/(225-176)#
= #-(432+125sqrt11)/49=-17.277#
