If sin⁻¹x + sin⁻¹y + sin⁻¹z = π, then x√(1−x²) + y√(1−y²) + z√(1−z²) equals:
If sin⁻¹x + sin⁻¹y + sin⁻¹z = π, then x√(1−x²) + y√(1−y²) + z√(1−z²) equals:
- A. 2xyz
- B. xyz
- C. 0
- D. 1
Answer: A) 2xyz
Explanation: Let sin⁻¹x = A, sin⁻¹y = B, sin⁻¹z = C. A+B+C=π. Then sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C. Here sin 2A = 2 sin A cos A = 2x√(1−x²). So sum = 2[x√(1−x²)+...] = 4xyz. Thus x√(1−x²)+y√(1−y²)+z√(1−z²) = 2xyz.
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