(b + c) tan(A/2) + (c + a) tan(B/2) + (a + b) tan(C/2) = 4R(cos A + cos B + cos C) = 4(R + r)

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Solution

$(b + c) t a n (A / 2)$
$= 2 R (s i n B + s i n C) t a n (A / 2)$
$= 2 R \cdot 2 s i n \frac{B + C}{2} \cdot c o s \frac{B - C}{2} \cdot s i n \frac{s i n (A / 2)}{c o s (A / 2)}$
$= 2 R \cdot 2 c o s \frac{B - C}{2} \cdot c o s \frac{B + C}{2}$
$= 2 R (c o s B + c o s C)$
$∴$ $L H S = 4 R (c o s A + c o s B + c o s C)$
$= 4 R (1 + \frac{r}{R})$ = $4 (R + r)$

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