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 -...

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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.` **