First slide
Properties of ITF
Question

If sin1a+sin1b+sin1c=π, then a1a2+b1b2+c1c2=

Moderate
Solution

Let A=sin1a,B=sin1b, and C=sin1c

we get A+B+C=π

a1a2+b1b2+c1c2=12(sin2A+sin2B+sin2C)=12[4sinAsinBsinC]=2abc

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