If A,B,C are the angles off `DeltaABC`, then `cotA*cotB+cotB*cotC+cotC*cotA`=

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If A,B ,C are the angle of ` Delta ABC ` then ` cot A .cot B + cot B . "" cot C + cot If A,B ,C are the angle of ` Delta ABC ` then ` cot A .cot B + cot B . "" cot C + cot
Prove that cotB cotC + cotC cotA + cotA cotB = 1 Prove that cotB cotC + cotC cotA + cotA cotB = 1
In `Delta ABC, (CotB + CotC) (CotC-CotA) (CotA+CotB) =` In `Delta ABC, (CotB + CotC) (CotC-CotA) (CotA+CotB) =`
If for `DeltaABC`, `cotA cotB cotC gt0` then the triangle is If for `DeltaABC`, `cotA cotB cotC gt0` then the triangle is
Trigonometry Formula | CotA CotB CotC for Any Triangle ABC | Proof of Cotangent | AMBPi73 Trigonometry Formula | CotA CotB CotC for Any Triangle ABC | Proof of Cotangent | AMBPi73
cotB.CotC+cotc.Cota+cota.Cotb equal to 1 @Mathby1069 cotB.CotC+cotc.Cota+cota.Cotb equal to 1 @Mathby1069
Show that cot(A-B) = (1+ cotA cotB)/(cotB -cotA). if cotA =½, cotB =2 and cotC =3 then cot(A-B-C)=3. Show that cot(A-B) = (1+ cotA cotB)/(cotB -cotA). if cotA =½, cotB =2 and cotC =3 then cot(A-B-C)=3.
If `A+B+C=180^@` then the value of `(cotB+cotC)(cotC+cotA)(cotA+cotB)` will be If `A+B+C=180^@` then the value of `(cotB+cotC)(cotC+cotA)(cotA+cotB)` will be