If A + B + C = 180^o, then the value of (cot | vedclass

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(B) Given $A + B + C = 180^o.$We know that $\cot B + \cot C = \frac{\sin C \cos B + \sin B \cos C}{\sin B \sin C} = \frac{\sin(B + C)}{\sin B \sin C}.

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$\csc A \csc B \csc C$

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(B) Given $A + B + C = 180^o.$We know that $\cot B + \cot C = \frac{\sin C \cos B + \sin B \cos C}{\sin B \sin C} = \frac{\sin(B + C)}{\sin B \sin C}.$Since $B + C = 180^o - A,$ we have $\sin(B + C) = \sin(180^o - A) = \sin A.$Thus,$\cot B + \cot C = \frac{\sin A}{\sin B \sin C}.$Similarly,$\cot C + \cot A = \frac{\sin B}{\sin C \sin A}$ and $\cot A + \cot B = \frac{\sin C}{\sin A \sin B}.$Multiplying these three expressions:$(\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B) = \left( \frac{\sin A}{\sin B \sin C} \right) \left( \frac{\sin B}{\sin C \sin A} \right) \left( \frac{\sin C}{\sin A \sin B} \right)$$= \frac{\sin A \sin B \sin C}{(\sin A \sin B \sin C)^2} = \frac{1}{\sin A \sin B \sin C} = \csc A \csc B \csc C.$

If $A + B + C = 180^o,$ then the value of $(\cot B + \cot C)(\cot C + \cot A)(\cot A + \cot B)$ is

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