Question

If sin⁻¹x+sin⁻¹y+sin⁻¹z= π, then prov that x√1-x²+ y√1-y²+z√1-z²=2xyz

Answer 1

Answer:

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

Step-by-step explanation:

sin⁻¹x+sin⁻¹y+sin⁻¹z= π

sin⁻¹x = A =>x = sinA    => √(1 - x²) = CosA

sin⁻¹y = B => y = sinB   => √(1 - y²) = CosB

sin⁻¹z = C => z = sinC    => √(1 - z²) = CosC

A + B + C = π  

to be proved

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

sinAcosA + sinBcosB +sinCcosC = 2sinAsinBsinC  

multiplying by 2 both sides

2sinAcosA + 2sinBcosB +2sinCcosC = 4sinAsinBsinC  

sin2A + sin2B + sin2C = 4sinAsinBsinC.

LHS  

=　sin2A + sin2B + sin2C  

= 2sin(A + B)cos(A - B) + sin[2π- 2A - 2B]  

= 2sin(A + B)cos(A - B) - sin(2A + 2B)  

= 2sin(A + B)cos(A - B) - 2sin(A + B)cos(A + B)  

= 2sin(A + B)[cos(A - B) - cos(A + B)]  

= 2sin(A + B)⋅2[(-1)(-1) sinAsinB]  

= 4sin(π - C)⋅sinAsinB  

= 4sinAsinBsinC

=RHS