In an acute angled triangle,cot B cot C + cot | vedclass

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(C) In $	riangle ABC$,the sum of angles is $A + B + C = 180^{\circ}$.$A + B = 180^{\circ} - C$Taking $\cot$ on both sides:$\cot(A + B) = \cot(180^{\c

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(C) In $	riangle ABC$,the sum of angles is $A + B + C = 180^{\circ}$.$A + B = 180^{\circ} - C$Taking $\cot$ on both sides:$\cot(A + B) = \cot(180^{\circ} - C)$Using the formula $\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$ and $\cot(180^{\circ} - C) = -\cot C$:$\frac{\cot A \cot B - 1}{\cot A + \cot B} = -\cot C$$\cot A \cot B - 1 = -\cot C(\cot A + \cot B)$$\cot A \cot B - 1 = -\cot C \cot A - \cot C \cot B$Rearranging the terms:$\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$

In an acute angled triangle,$\cot B \cot C + \cot A \cot C + \cot A \cot B$ is equal to

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