If {sin ^{ - 1}}a + {sin ^{ - 1}}b + {sin ^{ | vedclass

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(A) Let ${\sin ^{ - 1}}a = A, {\sin ^{ - 1}}b = B, {\sin ^{ - 1}}c = C$. Then $a = \sin A, b = \sin B, c = \sin C$ and $A + B + C = \pi$. We know that

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$2abc$

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(A) Let ${\sin ^{ - 1}}a = A, {\sin ^{ - 1}}b = B, {\sin ^{ - 1}}c = C$. Then $a = \sin A, b = \sin B, c = \sin C$ and $A + B + C = \pi$. We know that in a triangle where $A + B + C = \pi$,the identity $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$ holds. This can be written as $2\sin A \cos A + 2\sin B \cos B + 2\sin C \cos C = 4\sin A \sin B \sin C$. Dividing both sides by $2$,we get $\sin A \cos A + \sin B \cos B + \sin C \cos C = 2\sin A \sin B \sin C$. Since $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - a^2}$,$\cos B = \sqrt{1 - b^2}$,and $\cos C = \sqrt{1 - c^2}$,we substitute these values: $a\sqrt{1 - a^2} + b\sqrt{1 - b^2} + c\sqrt{1 - c^2} = 2abc$.

If ${\sin ^{ - 1}}a + {\sin ^{ - 1}}b + {\sin ^{ - 1}}c = \pi ,$ then the value of $a\sqrt {1 - {a^2}} + b\sqrt {1 - {b^2}} + c\sqrt {1 - {c^2}}$ is

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