If $$ABCD$$   is a convex quadrilateral such that $$4\sec A + 5 = 0$$    then the quadratic equation whose roots are $$\tan A$$  and $${\text{cosec}}\,A$$  is

Solution :
$$\sec A = - \frac{5}{4}.\,{\text{So, }}\frac{\pi }{2} < A < \pi .\,{\text{Hence,}}\tan A = - \frac{3}{4}\,{\text{and cosec}}\,A = \frac{5}{3}.$$
∴ the required equation is $${x^2} - \left( { - \frac{3}{4} + \frac{5}{3}} \right)x + \left( { - \frac{3}{4}} \right) \times \frac{5}{3} = 0.$$