If A+B+C=180∘ then sin2⁡A2+sin2⁡D2+sin2⁡C2=

=sin2⁡A2+1−cos2⁡B2+sin2⁡C2=1−cos2⁡B2−sin2⁡A2+sin2⁡C2=1−cos⁡A+B2cos⁡A−B2+sin2⁡C2=1−sin⁡C2cos⁡A−B2−cos⁡A+B2=1−2sin⁡A2sin⁡B2sin⁡C2

If A+B+C=180∘ then sin2⁡A2+sin2⁡D2+sin2⁡C2=

=sin2⁡A2+1−cos2⁡B2+sin2⁡C2=1−cos2⁡B2−sin2⁡A2+sin2⁡C2=1−cos⁡A+B2cos⁡A−B2+sin2⁡C2=1−sin⁡C2cos⁡A−B2−cos⁡A+B2=1−2sin⁡A2sin⁡B2sin⁡C2

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

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$1 - 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

$1 - 2 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$

$1 - 2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$

$1 - 2 \sin \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$

answer is A.

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Detailed Solution

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