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  4. If the roots of the equation x2+a x+b=0 are cot 60° and cot 75°, then (b-a) equals

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Q. If the roots of the equation $x^2+a x+b=0$ are $\cot 60^{\circ}$ and $\cot 75^{\circ}$, then $(b-a)$ equals

Complex Numbers and Quadratic Equations

Solution:

$- a =\cot 60^{\circ}+\cot 75^{\circ}=\frac{1}{\sqrt{3}}+2-\sqrt{3}$
$b =\cot 60^{\circ} \cdot \cot 75^{\circ}=\frac{2-\sqrt{3}}{\sqrt{3}}=\frac{2}{\sqrt{3}}-1 $
$\therefore( b - a )=\frac{2}{\sqrt{3}}-1+\frac{1}{\sqrt{3}}+2-\sqrt{3}=\sqrt{3}-1+2-\sqrt{3}=1$