<s>[INST] If $\sin \phi$ and $\tan \phi$ are the roots of the equation $ax^2+bx+c=0$, compute $b^2-c^2$ 
If $\sin \phi$ and $\tan \phi$ are the roots of the equation $ax^2+bx+c=0$. Then $(b^2-c^2) = $
$\bf{Options::}$ $(a)\;\; 4ac\;\;\;\;\;\;(b)\;\; a^2\;\;\;\;\;\;(c)\;\; 4bc\;\;\;\;\;\;(d)\;\; 4ab$

$\bf{My\; Try::}$ Using $\displaystyle \sin \phi+\tan \phi  = -\frac{b}{a}$ and $\displaystyle \sin \phi \cdot \tan \phi = \frac{c}{a}$.
Now $(b^2-c^2) = a^2(\sin \phi+\tan \phi)^2-a^2(\sin \phi \cdot \tan \phi)^2 = a^2 \left\{\sin^2 \phi+\tan^2 \phi+\tan \phi \cdot \sin \phi\right\}$
Now How can I solve after that
Thanks
 [/INST] Use $(\sin\phi)+(\tan\phi)=-b/a$, $(\sin\phi)(\tan\phi)=c/a$.
$$\begin{align}
(b^2-c^2)
&=a^2(\sin^2\phi+\tan^2\phi+2\sin\phi\tan\phi-\sin^2\phi\tan^2\phi)\\
&=a^2(\sin^2\phi+\tan^2\phi(1-\sin^2\phi)+2\sin\phi\tan\phi)\\
&=a^2(2\sin^2\phi+2\sin\phi\tan\phi)\\
&=2a^2(\sin\phi(\sin\phi+\tan\phi))\\
&=-2ab\sin\phi\end{align}$$
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