A) \[2\text{ }sin\text{ }A\text{ }sin\text{ }B\text{ }sin\text{ }C\]
B) \[2\text{ }cos\text{ }A\text{ }cos\,B\text{ }cos\text{ }C\]
C) \[4\text{ }sin\text{ }A\text{ }sin\text{ }B\text{ }sin\text{ }C\]
D) \[4\text{ }cos\text{ }A\text{ }cos\text{ B }cos\text{ }C\]
Correct Answer: D
Solution :
Given, \[A+B+C=\frac{\pi }{2}\] \[sin\text{ }2A+sin\text{ }2B+sin\text{ }2C\] \[=(sin\text{ }2A+sin\text{ }2B)+sin\text{ }2C\] \[=2\sin \left( \frac{2A+2B}{2} \right)\cos \left( \frac{2A-2B}{2} \right)+\sin 2C\] \[=2\sin (A+B)\cos (A-B)+\sin 2C\] \[=2\cos C.\cos (A-B)+2\sin C\cos C\] \[\left[ \begin{align} & \because A+B+C=\pi /2 \\ & A+B=\pi /2-C \\ \end{align} \right]\] \[=2\cos C[\cos (A-B)+\sin C]\] \[=2\cos C[\cos (A-B)+\cos (A+B)\] \[\left[ \because C=\frac{\pi }{2}-(A+B) \right]\] \[=2\cos C[2\cos A\cos B]\] \[=4\cos A\cos B\cos C\]
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