A+B+C =π/2 then
tan A tan B + tan B tan C + tan C tan A=
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tan A tan B + tan B tan C + tan C tan A=
tan( A+B) +tan (B+C) + tan (C+A)=
tan[(A+B)+(B+C)+(C+A)]=
tan[2(A+B+C)]=
tan(2×pai/2)=
tan(22/7).
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