Question
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 $$\cos C,\tan D$$ is
A.
$$39{x^2} - 16x - 48 = 0$$
B.
$$39{x^2} + 88x + 48 = 0$$
C.
$$39{x^2} - 88x + 48 = 0$$
D.
None of these
Answer :
$$39{x^2} - 16x - 48 = 0$$
Solution :
In a convex quadrilateral no angle is greater than $${180^ \circ }$$
$$\eqalign{
& {\text{Here, }}\tan A = \frac{5}{{12}}.\,{\text{So, }}0 < A < \frac{\pi }{2}\,{\text{and }}\frac{\pi }{2} < C < \pi \,\,\left( {\because \,\,A + C = {{180}^ \circ }} \right) \cr
& \therefore \,\,\tan \left( {\pi - C} \right) = \frac{5}{{12}},\,{\text{i}}{\text{.e}}{\text{., }}\tan C = - \frac{5}{{12}}\,\,\,\,\,\,\therefore \,\,\cos C = - \frac{{12}}{{13}}. \cr
& {\text{Also, }}\cos B = - \frac{3}{5}.\,{\text{So, }}\frac{\pi }{2} < B < \pi \,\,{\text{and }}0 < D < \frac{\pi }{2}\,\,\,\,\,\,\left( {\because \,\,B + D = {{180}^ \circ }} \right) \cr
& \therefore \,\,\cos \left( {\pi - D} \right) = - \frac{3}{5},\,{\text{i}}{\text{.e}}{\text{., }}\cos D = \frac{3}{5}\,\,\,\,\,\,\therefore \,\,\tan D = \frac{4}{3}. \cr} $$
∴ the required equation is $${x^2} - \left( { - \frac{{12}}{{13}} + \frac{4}{3}} \right)x + \left( { - \frac{{12}}{{13}}} \right) \cdot \frac{4}{3} = 0.$$