In any triangle ABC,a cot A + b cot B + c cot | vedclass

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(C) We know that $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these in the expression:$a \cot A + b \cot B + c \cot C = 2R \sin A

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$2(r + R)$

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(C) We know that $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these in the expression:$a \cot A + b \cot B + c \cot C = 2R \sin A \cot A + 2R \sin B \cot B + 2R \sin C \cot C$$= 2R (\cos A + \cos B + \cos C)$Using the identity $\cos A + \cos B + \cos C = 1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ and $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$,we get:$= 2R \left( 1 + \frac{r}{R} \right)$$= 2(R + r)$.

In any triangle $ABC$,$a \cot A + b \cot B + c \cot C = $

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