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Q. If $ABCD$ is a cyclic quadrilateral such that $3 \tan A +4=0$ and $13 \cos B +5=0$, then a quadratic equation whose roots are $\tan C$ and $\tan D$, is

Complex Numbers and Quadratic Equations

Solution:

$ \tan A =-\frac{4}{3} ; \because A + C =\pi \Rightarrow \tan C =\frac{4}{3}$ $\cos B =\frac{-5}{12} ;$ and $B + D =\pi \Rightarrow \cos D =\frac{5}{13} \Rightarrow \tan D =\frac{12}{5}$
$\therefore$ Equation are $\left(x-\frac{4}{3}\right)\left(x-\frac{12}{5}\right)=0$