In $Δ A B C$ , a, b, c are the lengths of its sides and A, B, C are the angles of triangle ABC. The correct relation is

(b - c) s i n (\frac{B - C}{2}) = a c o s (\frac{A}{2})

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(b - c) c o s (\frac{A}{2}) = a s i n (\frac{B - C}{2})

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(b + c) s i n (\frac{B + C}{2}) = a c o s (\frac{A}{2})

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(b - c) c o s (\frac{A}{2}) = 2 a s i n (\frac{B + C}{2})

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Solution

The correct option is B $(b - c) c o s (\frac{A}{2}) = a s i n (\frac{B - C}{2})$
In $△ A B C$
a, b, c are the length of its side and A, B, C are the angle of triangle ABC
$\Rightarrow \frac{b - c}{a} = \frac{sin B - sin C}{sin A} = \frac{2 cos (\frac{B + C}{2}) sin (\frac{B - C}{2})}{2 sin A / 2 . cos A / 2} = \frac{sin (\frac{B - C}{2})}{cos A / 2} [∵ cos \frac{B + C}{2} = sin A / 2]$
$\Rightarrow∴ (b - c) cos A / 2 = a sin (\frac{B - C}{2})$
$∴$ option B is correct.

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