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If $\cot \alpha$ and $\cot \beta$ are the roots of the equation $x^{2}+b x+c=0$ with $b \neq 0$, then the value of $\cot (\alpha+\beta)$
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The correct answer is:
$\frac{1-c}{b}$
Given equation, $\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}=0$
Roots are $\cot \alpha, \cot \beta$
Sum of roots $=\cot \alpha+\cot \beta=-b$
Product of roots $=\cot \alpha, \cot \beta=c$
$\cot (\alpha+\beta) \frac{\cot \alpha \cdot \cot \beta-1}{\cot \beta+\cot \alpha}=\frac{\mathrm{c}-1}{-\mathrm{b}}=\frac{1-\mathrm{c}}{\mathrm{b}}$
Roots are $\cot \alpha, \cot \beta$
Sum of roots $=\cot \alpha+\cot \beta=-b$
Product of roots $=\cot \alpha, \cot \beta=c$
$\cot (\alpha+\beta) \frac{\cot \alpha \cdot \cot \beta-1}{\cot \beta+\cot \alpha}=\frac{\mathrm{c}-1}{-\mathrm{b}}=\frac{1-\mathrm{c}}{\mathrm{b}}$
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