If ABCD is a cyclic quadrilateral such that 12 $\tan A -5 = 0$ and 5 $\cos B + 3 = 0,$ then the quadratic equation whose roots are cos C and tan D, is

 

Given that ABCD is a cyclic quadrilateral

$\therefore A+C=\pi \text{ and }B+D=\pi$

Now,

$\begin{array}{l}12\tan A-5=0\text{ and }5\cos B+3=0\\ \Rightarrow \tan A=\frac{5}{12}\text{ and }\cos B=-\frac{3}{5}\\ \Rightarrow \tan C=-\frac{5}{12}\text{ and }\cos D=\frac{3}{5}[\because A=\pi -C\text{ and }B=\pi -D]\\ \Rightarrow \cos C=-\frac{12}{13}\text{ and }\tan D=\frac{4}{3}\end{array}$

$\left[\begin{array}{l}\tan A>0\therefore 0<A<\frac{\pi }{2}\Rightarrow \frac{\pi }{2}<C<\pi \\ \cos B<0\therefore \frac{\pi }{2}<B<\pi \Rightarrow 0<D<\frac{\pi }{2}\end{array}\right]$

The equation having cos C and tan D as its roots is

$\begin{array}{l}{x}^2-x(\cos C+\tan D)+\cos C\tan D=0\\ \text{ or, }{x}^2-x\left(-\frac{12}{13}+\frac{4}{3}\right)+\left(-\frac{12}{13}\times \frac{4}{3}\right)=0\\ \text{ or, }39x^2-16x-48=0\end{array}$