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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
In $\triangle A B C, \frac{\cos \frac{1}{2}(B-C)}{\sin \frac{1}{2} A}=$
  • A
    $\frac{ b - c }{ a }$
  • ✓
    $\frac{ b + c }{ a }$
  • C
    $\frac{ a }{ b - c }$
  • D
    $\frac{a}{b+c}$
✓

Answer

Correct option: B.
$\frac{ b + c }{ a }$
(B) $\frac{\cos \frac{ B - C }{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{ B + C }{2} \cos \frac{B- C }{2}}{\sin \frac{B+ C }{2} \sin \frac{A}{2}}$
$=\frac{2 \sin \frac{B+ C }{2} \cos \frac{B- C }{2}}{2 \sin \left(\frac{\pi}{2}-\frac{ A }{2}\right) \sin \frac{ A }{2}}$
$=\frac{\sin B+\sin C}{\sin A}=\frac{b+c}{a}$

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