• # question_answer In a triangle $ABC$ the value of $\angle A$ is given by $5\cos A+3=0$, then the equation  whose roots are $\sin A$ and $\tan A$ will be [Roorkee 1972] A) $15{{x}^{2}}-8x+16=0$ B) $15{{x}^{2}}+8x-16=0$ C) $15{{x}^{2}}-8\sqrt{2}x+16=0$ D) $15{{x}^{2}}-8x-16=0$

Given that $5\cos A+3=0$or $\cos A=-\frac{3}{5}$ Let $\alpha =\sin A$and $\beta =\tan A$, then the sum of roots $=\alpha +\beta =\sin A+\tan A$ $=\sin A+\frac{\sin A}{\cos A}=\frac{\sin A}{\cos A}(1+\cos A)$ $=\frac{\sqrt{1-9/25}}{-3/5}\left( 1-\frac{3}{5} \right)=\frac{4}{-5}.\frac{5}{3}.\frac{2}{5}=\frac{8}{-15}$ and product of roots $\alpha .\beta =\sin A\tan A=\frac{{{\sin }^{2}}A}{\cos A}$ $=\frac{16/25}{-3/5}=-\frac{16}{25}\times \frac{5}{3}=-\frac{16}{15}$ Thus required equation  is ${{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0$ Þ ${{x}^{2}}+\frac{8x}{15}-\frac{16}{15}=0$ Þ $15{{x}^{2}}+8x-16=0$