If cot x cot y = a and x+y = frac{pi}{6},then | vedclass

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(B) Given $x+y = \frac{\pi}{6}$.Taking $\cot$ on both sides,$\cot(x+y) = \cot(\frac{\pi}{6}) = \sqrt{3}$.Using the formula $\cot(x+y) = \frac{\cot x \

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$\sqrt{3} t^2+(1-a) t+a \sqrt{3}=0$

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(B) Given $x+y = \frac{\pi}{6}$.Taking $\cot$ on both sides,$\cot(x+y) = \cot(\frac{\pi}{6}) = \sqrt{3}$.Using the formula $\cot(x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}$,we get $\frac{a-1}{\cot x + \cot y} = \sqrt{3}$.Therefore,$\cot x + \cot y = \frac{a-1}{\sqrt{3}}$.The quadratic equation with roots $\cot x$ and $\cot y$ is given by $t^2 - (\cot x + \cot y)t + (\cot x \cot y) = 0$.Substituting the values,$t^2 - \frac{a-1}{\sqrt{3}}t + a = 0$.Multiplying by $\sqrt{3}$,we get $\sqrt{3}t^2 - (a-1)t + a\sqrt{3} = 0$,which is $\sqrt{3}t^2 + (1-a)t + a\sqrt{3} = 0$.

If $\cot x \cot y = a$ and $x+y = \frac{\pi}{6}$,then the quadratic equation whose roots are $\cot x$ and $\cot y$ is

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