If sin−1a+sin−1b+sin−1c=π, then a1−a2+b1−b2+c1−c2=
a+b+c
a2b2c2
2abc
4abc
Let A=sin−1a,B=sin−1b, and C=sin−1c
we get A+B+C=π
a1−a2+b1−b2+c1−c2=12(sin2A+sin2B+sin2C)=12[4sinAsinBsinC]=2abc