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Question
If sin−1 a+sin−1 b+sin−1 c=π,then the value of a√(1−a2)+b√(1−b2)+c√(1−c2) will be
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Solution
The correct option is A 2abc
Let sin−1 a=A,sin−1 b=B,sin−1 c=C∴sin A=a,sin B=b,sin C=cand A+B+C=π,thensin2A+sin2B+sin2C=4 sin A sin B sin C........(i)⇒sin A cos A+sin B cos B+sin C cosC=2sin A sin B sin C⇒sin A√1−sin2 A+sin B√(1−sin2 B)+sin C√(1−sin2 C)=2 sin A sin B sin C. .....(ii)⇒a√(1−a2)+b√(1−b2)+c√(1−c)2=2abc,While sin−1a+sin−1b+sin−1c=π.
Let sin−1 a=A,sin−1 b=B,sin−1 c=C∴sin A=a,sin B=b,sin C=cand A+B+C=π,thensin2A+sin2B+sin2C=4 sin A sin B sin C........(i)⇒sin A cos A+sin B cos B+sin C cosC=2sin A sin B sin C⇒sin A√1−sin2 A+sin B√(1−sin2 B)+sin C√(1−sin2 C)=2 sin A sin B sin C. .....(ii)⇒a√(1−a2)+b√(1−b2)+c√(1−c)2=2abc,While sin−1a+sin−1b+sin−1c=π.
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