Question

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

• A. $8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$
• B. $8\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$
• 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 1

Answer: (B)

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

Answer 2

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