If cos A = cos B cos C and A + B + C = pi, th | vedclass

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(D) Given: $\cos A = \cos B \cos C$ and $A + B + C = \pi$.From the triangle property,$B + C = \pi - A$.Taking cosine on both sides,$\cos(B + C) = \cos

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$\frac{1}{2}$

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(D) Given: $\cos A = \cos B \cos C$ and $A + B + C = \pi$.From the triangle property,$B + C = \pi - A$.Taking cosine on both sides,$\cos(B + C) = \cos(\pi - A)$.Using the identity $\cos(B + C) = \cos B \cos C - \sin B \sin C$ and $\cos(\pi - A) = -\cos A$,we get:$\cos B \cos C - \sin B \sin C = -\cos A$.Substitute the given value $\cos A = \cos B \cos C$ into the equation:$\cos B \cos C - \sin B \sin C = -(\cos B \cos C)$.Rearranging the terms,we get:$2 \cos B \cos C = \sin B \sin C$.Dividing both sides by $2 \sin B \sin C$,we obtain:$\frac{\cos B \cos C}{\sin B \sin C} = \frac{1}{2}$.Therefore,$\cot B \cot C = \frac{1}{2}$.

If $\cos A = \cos B \cos C$ and $A + B + C = \pi,$ then the value of $\cot B \cot C$ is

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