Question
If sin-1a + sin-1b+ sin-1c = π , then the value of { a√(1-a2) + b√(1-b2) + c√(1-c2)}
Joshi sir comment
let sin-1a = A , sin-1b = B and sin-1c= C
so { a√(1-a2) + b√(1-b2) + c√(1-c2)} = sinAcosA + sinBcosB + sinCcosC = 1/2 (sin2A+sin2B+sin2C)
= 1/2 (4sinAsinBsinC)
= 2abc