If A, B and C are the angles of a \[\Delta ABC,\] then following is equal to \[\left( \frac{B+C}{2} \right)\] [SSC (CGL) 2015] |
A) \[\text{cosec}\frac{A}{2}\]
B) \[\cos \frac{A}{2}\]
C) \[\sec \frac{A}{2}\]
D) \[\sec \frac{B}{2}\]
Correct Answer: B
Solution :
A, B and C are angles of a triangle. |
\[\therefore \] \[\angle A+\angle B+\angle C=180{}^\circ \] |
\[\sin \left( \frac{B+C}{2} \right)=\sin \left( \frac{A+B+C}{2}-\frac{A}{2} \right)\] |
\[=\sin \left( 90{}^\circ -\frac{A}{2} \right)=\cos \frac{A}{2}\] |
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