A) \[2abc\]
B) \[abc\]
C) \[\frac{1}{2}abc\]
D) \[\frac{1}{3}abc\]
Correct Answer: A
Solution :
Let \[{{\sin }^{-1}}a=A,\] \[{{\sin }^{-1}}b=B,\] \[{{\sin }^{-1}}c=C\] \[\therefore \sin A=a,\sin B=b,\sin C=c\] and \[A+B+C=\pi ,\]then \[\sin 2A+\sin 2B+\sin 2C\] \[=4\sin A\,\,\sin B\,\,\sin C\] ?..(i) Þ \[\sin A\,\cos A\,+\sin B\,\cos B+\sin \,C\,\cos C\] = \[2\sin A\sin B\sin C\] Þ \[\sin A\sqrt{(1-{{\sin }^{2}}A)}+\sin B\sqrt{(1-{{\sin }^{2}}B)}+\sin C\sqrt{1-{{\sin }^{2}}C}\] \[=2\sin A\sin B\sin C.\] ??(ii) Þ \[a\sqrt{(1-{{a}^{2}})}+b\sqrt{(1-{{b}^{2}})}+c\sqrt{{{(1-c)}^{2}}}=2abc\], while \[{{\sin }^{-1}}a+{{\sin }^{-1}}b+{{\sin }^{-1}}c=\pi \].
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