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If R, r denote respectively the radius of the circumcircle and incircle of a triangle ABC, then
COLUMN-I COLUMN-II
A) $a \cos A + b \cos B + c \cos C$ p) $2 (R + r)$
B) $a \cot A + b \cot B + c \cot C$ q) $r$
C) $4 R \sin (A / 2) \sin (B / 2) \sin (C / 2)$ r) 4R $\sin A \sin B \sin C$
D) $\frac{a}{\sin A} + \frac{b}{\sin B} + \frac{c}{\sin C}$ s) $6 R$

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$A - s; B - r; C - q; D - p$

$A - r; B - p; C - q; D - s$

$A - q; B - r; C - s; D - p$

$A - r; B - s; C - q; D - p$

answer is B.

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Detailed Solution

$a \cos A + b \cos B + c \cos C = 2 R (\sin A \cos A + \sin B \cos B + \sin C \cos C) = R (\sin 2 A + \sin 2 B + \sin 2 C) = 4 R \sin A \sin B \sin C$
Next , $a \cot A + b \cot B + c \cot C$
= $2 R (\cos A + \cos B + \cos C)$
= 2R $[1 + 4 \sin (A / 2) \sin (B / 2) \sin (C / 2)]$
= 2(R + r)
As $4 R \sin (A / 2) \sin (B / 2) \sin (C / 2) = r$ (See the Text)

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