1.

Prove that: `cot^(-1)((ab+1)/(a-b))+cot^(-1)((bc+1)/(b-c))+cot^(-1)((ca+1)/(c-a))=0`.

Answer» We have,
LHS `=tan^(-1)(a-b)/(1+ab)+tan^(-1)(b-c)/(1+bc)+tan^(-1)(c-a)/(1+ca)`
`=(tan^(-1)a-tan^(-1)b)+(tan^(-1)b-tan^(-1)c)+(tan^(-1)c-tan^(-1)a)`
=0=RHS
`therefore` LHS=RHS.


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