If sin-1x+sin-1y+sin-1z=π then prove that x1-x2+y1-y2+z1-z2=2xyz.
Prove the given condition
Given, sin-1x+sin-1y+sin-1z=π
Let , sin-1x=A,sin-1y=B,sin-1z=C.
⇒x=sinA⇒y=sinB⇒z=sinC
Put values of x,y,z in sin-1x+sin-1y+sin-1z=π ,
⇒sin-1sinA+sin-1sinB+sin-1sinC=π⇒A+B+C=π
Substitute the values of x,y,z in x1-x2+y1-y2+z1-z2=2xyz
⇒LHS=sinA1-sin2A+sinB1-sin2B+sinC1-sin2C=sinAcosA+sinBcosB+sinCcosC...i∵sin2θ+cos2θ=1
Multiply and divide equation i by 2
⇒sinAcosA+sinBcosB+sinCcosC=122sinAcosA+2sinBcosB+2sinCcosC∵sin2θ=2sinθcosθ=12sin2A+sin2B+2sinCcosC∵sinA+sinB=2sinA+B2cosA-B2=122sinA+BcosA-B+2sinCcosC∵A+B=π-C=122sinCcosA-B+2sinCcosπ-A+B∵sinπ-C=sinC,cosπ-A+B=-cosA+B=sinCcosA-B-cosA+B∵cosA-cosB=-2sinA+B2sinA-B2=2sinCsinAsinB
Replacing back x,y,z in 2sinCsinAsinB,
⇒LHS=2xyz⇒LHS=RHS.
Hence proved.