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If the lines represented by $a x^2-b x y-y^2=0$ make angle $\alpha$ and $\beta$ with the positive direction of $X$-axis, then $\tan (\alpha+\beta)=$
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Verified Answer
The correct answer is:
$\frac{-b}{1+a}$
$\tan \alpha$ and $\tan \beta$ are roots of the $a x^2-b x y-y^2=0$
$$
\begin{aligned}
& \therefore \tan \alpha+\tan \beta=\frac{-(-\mathrm{b})}{-1}=-\mathrm{b} \text { and } \tan \alpha \tan \beta=\frac{\mathrm{a}}{(-1)}=-\mathrm{a} \\
& \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{-\mathrm{b}}{1-(-\mathrm{a})}=\frac{-\mathrm{b}}{1+\mathrm{a}}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \tan \alpha+\tan \beta=\frac{-(-\mathrm{b})}{-1}=-\mathrm{b} \text { and } \tan \alpha \tan \beta=\frac{\mathrm{a}}{(-1)}=-\mathrm{a} \\
& \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{-\mathrm{b}}{1-(-\mathrm{a})}=\frac{-\mathrm{b}}{1+\mathrm{a}}
\end{aligned}
$$
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