If A + B + C = 180^o , then 2A + 2B + 2C A + B + C - 1 = - Vidyadip

MCQ

If $A + B + C = {180^o},$ then $\frac{{\sin 2A + \sin 2B + \sin 2C}}{{\cos A + \cos B + \cos C - 1}} = $

A
$8\,\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
✓
$8\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
C
$8\,\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$
D
$8\,\cos \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

✓

Answer

Correct option: B.

$8\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

b
(b) Here ${D^r} = 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

and ${N^r} = 4\sin A\sin B\sin c$

$\therefore L.H.S. = \frac{{{N^r}}}{{{D^r}}}$

and $\sin A = 2\sin \frac{A}{2}\cos \frac{A}{2}$ .

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