If A+B+C=180° Prove that cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)

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Show that in a triangle ABC, cot(A/2) + cot(B/2) + cot(C/2) = cot(A/2)*cot(B/2)*cot(C/2)Show that in a triangle ABC, cot(A/2) + cot(B/2) + cot(C/2) = cot(A/2)*cot(B/2)*cot(C/2)
If `A+B+C=180^(@)`, then the value of `cot.(A)/(2)+cot.(B)/(2)+cot.(C)/(2)` will beIf `A+B+C=180^(@)`, then the value of `cot.(A)/(2)+cot.(B)/(2)+cot.(C)/(2)` will be
In any triangle ABC, prove that, (b^(2)-c^(2))cot A + (c^(2)-a^(2)) cot B + (a^(2)-b^(2)) cot C=...In any triangle ABC, prove that, (b^(2)-c^(2))cot A + (c^(2)-a^(2)) cot B + (a^(2)-b^(2)) cot C=...
`cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)``cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)`
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`cot(A/2)+cot(B/2)+cot(C/2)=``cot(A/2)+cot(B/2)+cot(C/2)=`
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