If A+B+C=180∘ then sin⁡2A+sin⁡2B+sin⁡2Ccos⁡A+cos⁡B+cos⁡C−1=

(sin⁡2A+sin⁡2B)+sin⁡2C=2sin⁡(A+B)cos⁡(A−B)+sin⁡2C=2sin⁡(π−C)cos⁡(A−B)+sin⁡2C=2sin⁡Ccos⁡(A−B)+2sin⁡Ccos⁡C=2sin⁡C[cos⁡(A−B)+cos⁡C]=2sin⁡C[cos⁡(A−B)+cos⁡{π−(A+B)}]=2sin⁡C[cos⁡(A−B)−cos⁡(A+B)]=2sin⁡C×2sin⁡Asin⁡B=4sin⁡Asin⁡Bsin⁡C(cos⁡A+cos⁡B)+cos⁡C−1=2cos⁡A+B2cos⁡A−B2+cos⁡C−1=2cos⁡π2−C2cos⁡A−B2+cos⁡C−1=2sin⁡C2cos⁡A−B2+1−2sin2⁡C2−1=2sin⁡C2cos⁡A−B2−2sin⁡2C2=2sin⁡C2cos⁡A−B2−sin⁡C2=2cos⁡C2cos⁡A−B2−sin⁡π2−A+B2=2sin⁡C2cos⁡A−B2−cos⁡A+B2=2sin⁡C22sin⁡A2sin⁡B2=4sin⁡A2sin⁡B2sin⁡C2sin⁡2A+sin⁡2B+sin⁡2Ccos⁡A+cos⁡B+cos⁡C−1=8cos⁡A/2cos⁡B/2cos⁡C/2

If A+B+C=180∘ then sin⁡2A+sin⁡2B+sin⁡2Ccos⁡A+cos⁡B+cos⁡C−1=

(sin⁡2A+sin⁡2B)+sin⁡2C=2sin⁡(A+B)cos⁡(A−B)+sin⁡2C=2sin⁡(π−C)cos⁡(A−B)+sin⁡2C=2sin⁡Ccos⁡(A−B)+2sin⁡Ccos⁡C=2sin⁡C[cos⁡(A−B)+cos⁡C]=2sin⁡C[cos⁡(A−B)+cos⁡{π−(A+B)}]=2sin⁡C[cos⁡(A−B)−cos⁡(A+B)]=2sin⁡C×2sin⁡Asin⁡B=4sin⁡Asin⁡Bsin⁡C(cos⁡A+cos⁡B)+cos⁡C−1=2cos⁡A+B2cos⁡A−B2+cos⁡C−1=2cos⁡π2−C2cos⁡A−B2+cos⁡C−1=2sin⁡C2cos⁡A−B2+1−2sin2⁡C2−1=2sin⁡C2cos⁡A−B2−2sin⁡2C2=2sin⁡C2cos⁡A−B2−sin⁡C2=2cos⁡C2cos⁡A−B2−sin⁡π2−A+B2=2sin⁡C2cos⁡A−B2−cos⁡A+B2=2sin⁡C22sin⁡A2sin⁡B2=4sin⁡A2sin⁡B2sin⁡C2sin⁡2A+sin⁡2B+sin⁡2Ccos⁡A+cos⁡B+cos⁡C−1=8cos⁡A/2cos⁡B/2cos⁡C/2

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

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$4 \sin A / 2 \sin B / 2 \sin C / 2$

$8 \cos A / 2 \cos B / 2 \cos C / 2$

$1 + 4 \sin A / 2 \sin B / 2 \sin C / 2$

$4 \cos A / 2 \cos B / 2 \cos C / 2$

answer is C.

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Detailed Solution

$\begin{array}{l} (\sin 2 A + \sin 2 B) + \sin 2 C \\ = 2 \sin (A + B) \cos (A - B) + \sin 2 C \\ = 2 \sin (π - C) \cos (A - B) + \sin 2 C \\ = 2 \sin C \cos (A - B) + 2 \sin C \cos C \\ = 2 \sin C [\cos (A - B) + \cos C] \\ = 2 \sin C [\cos (A - B) + \cos {π - (A + B)}] \\ = 2 \sin C [\cos (A - B) - \cos (A + B)] \\ = 2 \sin C \times 2 \sin A \sin B \\ = 4 \sin A \sin B \sin C \end{array}$

$\begin{array}{l} (\cos A + \cos B) + \cos C - 1 \\ = 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2} + \cos C - 1 \\ = 2 \cos (\frac{π}{2} - \frac{C}{2}) \cos \frac{A - B}{2} + \cos C - 1 \\ = 2 \sin \frac{C}{2} \cos \frac{A - B}{2} + 1 - 2 \sin^{2} \frac{C}{2} - 1 \\ = 2 \sin \frac{C}{2} \cos \frac{A - B}{2} - 2 \sin^{2} \frac{C}{2} \\ = 2 \sin \frac{C}{2} [\cos \frac{A - B}{2} - \sin \frac{C}{2}] \\ = 2 \cos \frac{C}{2} [\cos \frac{A - B}{2} - \sin (\frac{π}{2} - \frac{A + B}{2})] \\ = 2 \sin \frac{C}{2} [\cos \frac{A - B}{2} - \cos \frac{A + B}{2}] \\ = 2 \sin \frac{C}{2} (2 \sin \frac{A}{2} \sin \frac{B}{2}) = 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \end{array}$