Question

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

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (D)

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