You need to consider that A,B,C are angles in triangle ABC, hence `A+B+C=pi =gt A+B = pi-C` .
Hence `ctg(A+B) = ctg(pi - C)`
`(ctgA*ctgB - 1)/(ctgA + ctgB) = 1/tan(pi-C)`
`(ctgA*ctgB - 1)/(ctgA + ctgB) = (1 + tanpi*tanC)/(tan pi - tanC)`
`(ctgA*ctgB - 1)/(ctgA + ctgB) = 1/(-tan C)`
`(ctgA*ctgB - 1)/(ctgA + ctgB) = - ctg C`
`(ctgA*ctgB - 1) = -ctg C(ctgA + ctgB)`
You need to opent the brackets such that:
`ctgA*ctgB - 1= -ctgA*ctgC - ctgC*ctgB`
You need to move the terms containing cotangent function to the left side such that:
`ctgA*ctgB + ctgA*ctgC + ctgB*ctgC - 1=0`
`ctgA*ctgB + ctgA*ctgC + ctgB*ctgC = 1`
Hence, the last line proves the identity `ctgA*ctgB + ctgA*ctgC + ctgB*ctgC = 1.`
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