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If the roots of the equation $x^{2}+p x+q=0$ are $\tan 19^{\circ}$ and $\tan 26^{\circ}$, then which one of the following is correct?
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Verified Answer
The correct answer is:
$q-p=1$
Given, $\tan 19^{\circ}$ and $\tan 26^{\circ}$ are roots of $x^{2}+p x+q=0$
$\therefore \tan 19^{\circ}+\tan 26^{\circ}=\frac{-p}{1}=-p$
$\left(\tan 19^{\circ}\right)\left(\tan 26^{\circ}\right)=\frac{q}{1}=q$
$\tan \left(19^{\circ}+26^{\circ}\right)=\frac{\tan 19^{\circ}+\tan 26^{\circ}}{1-\tan 19^{\circ} \tan 26^{\circ}}$
$\tan 45^{\circ}=\frac{-p}{1-q} \Rightarrow 1=\frac{-p}{1-q}$
$\Rightarrow 1-q=-p$
$\Rightarrow q-p=1$
$\therefore \tan 19^{\circ}+\tan 26^{\circ}=\frac{-p}{1}=-p$
$\left(\tan 19^{\circ}\right)\left(\tan 26^{\circ}\right)=\frac{q}{1}=q$
$\tan \left(19^{\circ}+26^{\circ}\right)=\frac{\tan 19^{\circ}+\tan 26^{\circ}}{1-\tan 19^{\circ} \tan 26^{\circ}}$
$\tan 45^{\circ}=\frac{-p}{1-q} \Rightarrow 1=\frac{-p}{1-q}$
$\Rightarrow 1-q=-p$
$\Rightarrow q-p=1$
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