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(B) Given that $	an 	heta$ and $\cot 	heta$ are roots of $ax^2 + bx + c = 0$.From the properties of quadratic equations,the sum of roots is $	an \

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$\sin 2	heta = -\frac{2c}{b}$

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(B) Given that $	an 	heta$ and $\cot 	heta$ are roots of $ax^2 + bx + c = 0$.From the properties of quadratic equations,the sum of roots is $	an 	heta + \cot 	heta = -\frac{b}{a}$ and the product of roots is $	an 	heta \cdot \cot 	heta = \frac{c}{a}$.Since $	an 	heta \cdot \cot 	heta = 1$,we have $\frac{c}{a} = 1$,which implies $c = a$.Now,$	an 	heta + \cot 	heta = \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} = \frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta} = \frac{1}{\sin 	heta \cos 	heta} = \frac{2}{\sin 2	heta}$.Equating this to the sum of roots: $\frac{2}{\sin 2	heta} = -\frac{b}{a}$.Since $a = c$,we substitute $a$ with $c$: $\frac{2}{\sin 2	heta} = -\frac{b}{c}$.Therefore,$\sin 2	heta = -\frac{2c}{b}$.

If $\tan \theta$ and $\cot \theta$ are two distinct roots of the equation $ax^2 + bx + c = 0$,$a \neq 0, b \neq 0$,then

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