First, find tan A and tan B.
#cos A = 3/5# --> #sin^2 A = 1 - 9/25 = 16/25# --> #cos A = +- 4/5#
#cos A = 4/5# because A is in Quadrant I
#tan A = sin A/(cos A) = (4/5)(5/3) = 4/3.#
#sin B = 5/13# --> #cos^2 B = 1 - 25/169 = 144/169# --> #sin B = +- 12/13.#
#sin B = 12/13# because B is in Quadrant I
#tan B = sin B/(cos B) = (5/13)(13/12) = 5/12#
Apply the trig identity:
#tan (A - B) = (tan A - tan B)/(1 - tan A.tan B)#
#tan A - tan B = 4/3 - 5/12 = 11/12#
#(1 - tan A.tan B) = 1 - 20/36 = 16/36 = 4/9#
#tan (A - B) = (11/12)(9/4) = 33/16#
