If `x , y , z` are natural numbers such that `cot^(-1)x+cot^(-1)y=cot^(-1)z` then the number of ordered triplets `(x , y , z)` that satisfy the equation is 0 (b) 1 (c) 2 (d) Infinite solutions

Explanation

`cot^-1x+cot^-1y = cot^-1z`

`=>tan^-1(1/x)+tan^-1(1/y) = tan^-1(1/z)`br>
`=>tan^-1((1/x+1/y)/(1-(1/x)(1/y))) = tan^-1(1/z)`

`=>tan^-1((x+y)/(xy-1)) = tan^-1(1/z)`

`=>xy/(xy-1) = 1/z`

`=>x = (-(1+yz))/(z-y)`

Let `y = n, z=n+1, then, x = n^2+n+1`

So, for different values of `n`, we can have different solutions.

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