Trigonometric Equations Question 241
Question: In $ \Delta ,ABC $ , $ \frac{\cos \frac{1}{2}(B-C)}{\sin \frac{1}{2}A}= $
[MP PET 1993; Roorkee 1973]
Options:
A) $ \frac{b-c}{a} $
B) $ \frac{b+c}{a} $
C) $ \frac{a}{b-c} $
D) $ \frac{a}{b+c} $
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Answer:
Correct Answer: B
Solution:
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$ \frac{\cos \frac{B-C}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{B+C}{2}\cos \frac{B-C}{2}}{\sin \frac{B+C}{2}\sin \frac{A}{2}}=\frac{\sin B+\sin C}{\sin A}=\frac{b+c}{a} $ .
BETA
