jika tan A = 5/12 dan cot B = 4/3, dengan A lancip dan B tumpul. Maka nilai cot A. sec B/sin A.sin B adalah.. a. 13 b. 14 c. 15 d. 16 e. 17 tolong bantu dengan

tan A = 5/12  --> tan A = y/x = 5/12
r = √(x²+ y²) = √169)= 13
sin A = y/r = 5/13
cot A = 12/5

cot B = - 4/3 ---> tan B = - 3/4
tan B = y/x = - 3/4
r = √(x² +y²) = √25 = 5
sin B =  y/r = 3/5
cos B = - x r = - 4/5
sec B = - 5/4

cot A . sec B/ (sin  A  sin B) =
= (12/5)(-5/4) / (5/13)(3/5)
= - 3 / (3/13)
= - 3 x 13/3
= - 13.