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)#?

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Answer 1 · 2018-03-24T05:34:31

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)#