QUESTION

2 tanA - 5 =0 gives tanA = 5/12

hence, sin A = 5/13 and cos A = 12/13

5 cosB + 3 =0 gives cos B = -3/5

hence, sin B = 4/5

ABCD is cyclic quad. hence, A+C = 180 and B+D = 180

hence, A, B, C and D are all less than 180^o. so, they all are either in 1st or in 2nd quadrant.

so, sin will always take positive values, but cos and tan can take negative values as well.

cos C + tan D = -cos A -tan B = -12/13+4/3 = 16/39

cos C. tan D = (-cos A).(-tan B) =(-12/13) .(-4/3) =16/13

thus, the required quadratic equation is:

x²-16/39 x + 16/13=0

39x²-16x+48=0