QUESTION

Asked by  | 26th Jun, 2008, 09:49: PM

Expert Answer:

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 180o. 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

Answered by  | 15th Jul, 2008, 01:07: PM

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