Question

If sin^–1x + sin^–1y + sin^–1z = π, prove that x√(1 - x²) + y√(1 - y²) + z√(1 - z²)  = 2xyz

Answer 1

Let sin^–1x = α, sin^–1y = β and sin^–1z = γ. Then sin α = x, sin β = y, sin γ = z

=> sin^–1x + sin^–1y + sin^–1z = π

=> α + β + γ = π

=> 2 sin (π – C) . cos (A – B) + sin 2C = 2 sin C . cos(A – B) + 2 sin C. cos C

=>  2sinC [cos(A – B) + cos C)] = 2 sin C[cos (A – B) + cos(π – (A + B))]

=> 2 sin C [cos (A – B) – cos (A + B)] = 2 sin C [2sin A. sin B] = 4 sin A sin B sin C

=> 2 sin A . cos A + 2 sin B . cos B + 2 sin C . cos C = 4 sin A . sin B . sin C

=> sin A√(1-sin²A) + sin B√(1-sin²B) + sin C√(1-sin²C) = 2sinAsinBsinC

=> x√(1-x²) + y√(1-y²) + z√(1-z²) = 2xyz