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


  • $\begingroup$ 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. $\endgroup$ – almagest Sep 16 '14 at 18:07
  • $\begingroup$ 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 $\alpha=\sin\phi,\beta=\tan\phi$. $\endgroup$ – almagest Sep 16 '14 at 18:10
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    $\begingroup$ 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] $\endgroup$ – almagest Sep 16 '14 at 18:17
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    $\begingroup$ 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). $\endgroup$ – almagest Sep 16 '14 at 18:21

Use $(\sin\phi)+(\tan\phi)=-b/a$, $(\sin\phi)(\tan\phi)=c/a$.

$$\begin{align} (b^2-c^2) &=a^2(\sin^2\phi+\tan^2\phi+2\sin\phi\tan\phi-\sin^2\phi\tan^2\phi)\\ &=a^2(\sin^2\phi+\tan^2\phi(1-\sin^2\phi)+2\sin\phi\tan\phi)\\ &=a^2(2\sin^2\phi+2\sin\phi\tan\phi)\\ &=2a^2(\sin\phi(\sin\phi+\tan\phi))\\ &=-2ab\sin\phi\end{align}$$

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

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