If A , B ,C, D are angles of a cyclic quadrilateral ABCD such that 12 tan A - 5 = 0 and 5 cos D+3 = 0, find the quadratic equation whose roots are tan C and cos B Please help it is urgent
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tanA=125 and cosB=5−3
For cyclic quadrilateral, A+C=π and B+D=π
Therefore, tanC=tan(π−A)=12−5 and cosD=cos(π−C)=53
⇒cosC=13−12 and tanD=34
Hence required quadratic equation is,
x2−(13−12+34)+(13−12.34)=0
⇒39x2−16x−48=0
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