In Delta ABC,a cot A + b cot B + c cot C = . | 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 \frac{\cos A}{\sin A} + 2R \sin B \frac{\cos B}{\sin B} + 2R \sin C \frac{\cos C}{\sin C}$$= 2R(\cos A + \cos B + \cos C) \cdots (i)$Using the identity $\cos A + \cos B + \cos C = 1 + 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$ and the relation $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$:$\cos A + \cos B + \cos C = 1 + \frac{r}{R} = \frac{R + r}{R}$Substituting this into equation $(i)$:$2R(\frac{R + r}{R}) = 2(R + r)$.

In $\Delta ABC$,$a \cot A + b \cot B + c \cot C = . . . $ (where $r$ is inradius and $R$ is circumradius.)

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