In a triangle ABC,suppose none of the angles | vedclass

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(B) Given that $A + B + C = \pi$ and $A, B, C 
eq \frac{n\pi}{2}$.We know the identity $\cot(B + C) = \frac{\cot B \cot C - 1}{\cot B + \cot C}$.Sinc

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$1$

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(B) Given that $A + B + C = \pi$ and $A, B, C 
eq \frac{n\pi}{2}$.We know the identity $\cot(B + C) = \frac{\cot B \cot C - 1}{\cot B + \cot C}$.Since $B + C = \pi - A$,we have $\cot(B + C) = \cot(\pi - A) = -\cot A$.Substituting this into the identity:$-\cot A = \frac{\cot B \cot C - 1}{\cot B + \cot C}$.Multiplying both sides by $(\cot B + \cot C)$:$-\cot A(\cot B + \cot C) = \cot B \cot C - 1$.$-\cot A \cot B - \cot A \cot C = \cot B \cot C - 1$.Rearranging the terms gives:$\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$.

In a $\triangle ABC$,suppose none of the angles are multiples of $\frac{\pi}{2}$,then what is the value of $\cot A \cot B + \cot B \cot C + \cot A \cot C$?

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