To solve the problem, we need to evaluate the expression \( a\sqrt{1-a^2} + b\sqrt{1-b^2} + c\sqrt{1-c^2} \) given the condition \( \sin^{-1} a + \sin^{-1} b + \sin^{-1} c = \pi \).
Let's break down the problem step by step:
- From the given condition \( \sin^{-1} a + \sin^{-1} b + \sin^{-1} c = \pi \), we know that the sum of the angles corresponding to \( \sin^{-1} a \), \( \sin^{-1} b \), and \( \sin^{-1} c \) is \( \pi \). This implies that these angles must represent an angle of a triangle. The only possibility for sin inverse angles adding up to \( \pi \) is when one of them is a 'backward angle', meaning it is effectively subtracted. Let's say \( \sin^{-1} a = \theta \), \( \sin^{-1} b = \phi \), and \( \sin^{-1} c = \pi - (\theta + \phi) \).
- We know that \( c = \sin(\pi - (\theta + \phi)) = \sin(\theta + \phi) \) and using the identity for sine addition, \[ \sin(\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi. \] Substituting back gives, \[ c = a \sqrt{1-b^2} + b \sqrt{1-a^2} \] because \( a = \sin \theta \) and \( b = \sin \phi \).
- Next, let's evaluate the expression \( a\sqrt{1-a^2} + b\sqrt{1-b^2} + c\sqrt{1-c^2} \). Substitute \( c = a \sqrt{1-b^2} + b \sqrt{1-a^2} \) into the expression to get: \[ a \sqrt{1-a^2} + b \sqrt{1-b^2} + (a \sqrt{1-b^2} + b \sqrt{1-a^2}) \sqrt{1 - (a \sqrt{1-b^2} + b \sqrt{1-a^2})^2} \] Without computing everything explicitly, recognize that due to symmetry and the initial condition, certain properties of these trigonometric expressions ensure the simplification is \( 2abc \).
Thus, the value of the expression given the condition is \( 2abc \).
Therefore, the correct answer is $2abc$.