$If{\tan }^2\theta ,{\sin }^2\theta aretherootsofa{x}^2+bx+c=0then{b}^2-c^2=$

$\begin{array}{l}If\alpha =Ta{n}^2\theta ,\beta ={\sin }^2\theta arerootsofa{x}^2+bx+c=0\\ {(\alpha +\beta )}^2={(\alpha -\beta )}^2+4\alpha \beta \\ \Rightarrow {\left(\frac{-b}{a}\right)}^2={({\sec }^2\theta -{\tan }^2\theta )}^2+4\left(\frac{c}{a}\right)\\ \Rightarrow \frac{b^2}{a^2}=1+\frac{4c}{a}\\ \Rightarrow b^2=a^2+4ac\\ \Rightarrow b^2-a^2=4ac\end{array}$