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Trigonometric Ratios of Standard Angles

In triangle ...

Question

In triangle $A B C$ $a, b, c$ are the length of its sides and $A, B, C$ are the angles of triangle $A B C$ . The correct relation is given by

(b - c) sin (\frac{B - C}{2}) = a cos \frac{A}{2}

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(b - c) cos (\frac{A}{2}) = a sin \frac{B - C}{2}

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(b + c) sin (\frac{B + C}{2}) = a cos \frac{A}{2}

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(b - c) cos (\frac{A}{2}) = a sin \frac{B + C}{2}

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Solution

The correct option is B $(b - c) cos (\frac{A}{2}) = a sin \frac{B - C}{2}$

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