ta có tanA/2.tanB/2+tanB/2.tanC/2+tanC/2.tanA/2=1
----> cotA/2+cotAB/2+cotC/2=cotA/2cotB/2cotC/2
do đó:
1/2(tan(A/2) + tan(B/2) + tan(C/2) + cot(A/2).cot(B/2).cot(C/2))=1/2(tan(A/2) + tan(B/2) + tan(C/2) + cotA/2+cotAB/2+cotC/2)=1/2(tanA/2+cotA/2+tanB/2+cotB/2+tanC/2+cotC/2)
=
vậy
1/2[tan(A/2) + tan(B/2) + tan(C/2) + cot(A/2).cot(B/2).cot(C/2)=1/sinA + 1/sinB + 1/sinC
sai thôi nhé