# 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

• It might be easier to start from a known relationship between $\sin\phi$ and $\tan\phi$ and then express that in terms of the coefficients. – almagest Sep 16 '14 at 18:07
• You have for example that $\sin^2\phi+\cos^2\phi=1$ which leads immediately to a relationship between $\sin\phi$ and $\tan\phi$. It is clearer if you put $\alpha=\sin\phi,\beta=\tan\phi$. – almagest Sep 16 '14 at 18:10
• Since the resulting relationship does not seem that pleasant, are you sure you have the question correct? [Always a good thing to ask if things seem to be getting unexpectedly difficult] – almagest Sep 16 '14 at 18:17
• If you try putting $\phi=1$ and $a=1$, work out the roots and compare $b^2-c^2$ to the four alternatives, they all seem to be wrong (unless my calculation is botched, which is entirely possible). – almagest Sep 16 '14 at 18:21

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}
• Isn't it $2a^2(\sin\phi(\sin\phi + \tan\phi))=-2ab\sin\phi$? – user26486 Sep 16 '14 at 18:45