Prove that:- cot(A+B+C)=(cotAcotBcotC-cotA-cotB-cotC)/(cotAcotB+cotBcotC+cotAcotC-1)

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Prove Cot A Cot B + Cot B Cot C + Cot A Cot C = 1 if A+B+C = π. Prove Cot A Cot B + Cot B Cot C + Cot A Cot C = 1 if A+B+C = π.
cot(a + B) = (cota*cotB - 1) / (cotB + cota) cot(a + B) = (cota*cotB - 1) / (cotB + cota)
cot(a - B) = (cotA*cotB + 1) / (cotB - cotA) cot(a - B) = (cotA*cotB + 1) / (cotB - cotA)
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If CotA+CotB+CotC=√3 ||then Prove that ∆ABC forms an equilateral Triangle || in Telugu || If CotA+CotB+CotC=√3 ||then Prove that ∆ABC forms an equilateral Triangle || in Telugu ||
#cate original voice #shorts #cate original voice #shorts
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