Prove that: if A+B+C=180° cot(A/2)×cot (B/2)×cot (C/2)=cot(A/2)+cot (B/2)+cot (C/2)?

1 Answer
Mar 24, 2018

Given A+B+C=180^@

=>A/2+B/2+C/2=90^@

=>cot(A/2+B/2)=cot(90^@-C/2)

=>(cot(A/2)cot(B/2)-1)/(cot(A/2)+cot(B/2))=tan(C/2)

=>(cot(A/2)cot(B/2)-1)/(cot(A/2)+cot(B/2))=1/cot(C/2)

=>cot(A/2)xxcot(B/2)xxcot(C/2)-cot(C/2))=cot(A/2)+cot(B/2)

=>cot(A/2)xxcot(B/2)xxcot(C/2)=cot(A/2)+cot(B/2)+cot(C/2)