If `x > y > z >0,` then find the value of `cot^(-1)(x y+1)/(x-y)+cot^(-1)(y z+1)/(z y-z)+cot^(-1)(z x+1)/(z-x)`

Explanation

When `x gt 0, cot^-1x = tan^-1(1/x)`

When `x lt 0, cot^-1x = pi+ tan^-1(1/x)`

It is given that, `x gt y gt zgt 0`.

Now, `cot^-1((xy+1)/(x-y)) + cot^-1((yz+1)/(y-z))+cot^-1((zx+1)/(z-x))`

`=tan^-1((x-y)/(1+xy))+tan^-1((y-z)/(1+yz))+pi+tan^-1((z-x)/(1+zx))...[As (z - x) lt 0]`

`=tan^-1x-tan^-1y+tan^-1y-tan^-1z+pi+tan^-1z-tan^-1x`

`=pi`

`:. cot^-1((xy+1)/(x-y)) + cot^-1((yz+1)/(y-z))+cot^-1((zx+1)/(z-x)) = pi`