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Question

If sin-1x+sin-1y+sin-1z=π then prove that x1-x2+y1-y2+z1-z2=2xyz.


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Solution

Prove the given condition

Given, sin-1x+sin-1y+sin-1z=π

Let , sin-1x=A,sin-1y=B,sin-1z=C.

x=sinAy=sinBz=sinC

Put values of x,y,z in sin-1x+sin-1y+sin-1z=π ,

sin-1sinA+sin-1sinB+sin-1sinC=πA+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...isin2θ+cos2θ=1

Multiply and divide equation i by 2

sinAcosA+sinBcosB+sinCcosC=122sinAcosA+2sinBcosB+2sinCcosCsin2θ=2sinθcosθ=12sin2A+sin2B+2sinCcosCsinA+sinB=2sinA+B2cosA-B2=122sinA+BcosA-B+2sinCcosCA+B=π-C=122sinCcosA-B+2sinCcosπ-A+Bsinπ-C=sinC,cosπ-A+B=-cosA+B=sinCcosA-B-cosA+BcosA-cosB=-2sinA+B2sinA-B2=2sinCsinAsinB

Replacing back x,y,z in 2sinCsinAsinB,

LHS=2xyzLHS=RHS.

Hence proved.


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