If A, B, C are the angles of triangle ABC,the | vedclass

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(B) Given that $A + B + C = \pi$. Since $A + B = \pi - C$,we take the cotangent on both sides: $\cot(A + B) = \cot(\pi - C)$. Using the identity $\cot

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$1$

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(B) Given that $A + B + C = \pi$. Since $A + B = \pi - C$,we take the cotangent on both sides: $\cot(A + B) = \cot(\pi - C)$. Using the identity $\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$ and $\cot(\pi - C) = -\cot C$,we get: $\frac{\cot A \cot B - 1}{\cot A + \cot B} = -\cot C$. Multiplying both sides by $(\cot A + \cot B)$,we obtain: $\cot A \cot B - 1 = -\cot A \cot C - \cot B \cot C$. Rearranging the terms,we get: $\cot A \cot B + \cot B \cot C + \cot A \cot C = 1$.

If $A, B, C$ are the angles of $\triangle ABC$,then $\cot A \cot B + \cot B \cot C + \cot A \cot C = $

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