Question

`cot^-1x + cot^-1(1+x)=pi/4`?

Answer 1

`x=-1, 2`

Explanation:

`cot^(-1)(x)+cot^(-1)(1+x)=pi/4`

Take `tan` of both sides:

`tan[cot^(-1)(x)+cot^(-1)(1+x)]=tan(pi/4)`

Expand the left side using the sum formula:

`tan(x+y) = [tan(x)+tan(y)]/[1-tan(x)tan(y)]`

So in this case:

`[tan (cot^(-1)(x))+tan(cot^(-1)(1+x))]/[1-tan(cot^(-1)(x))*tan(cot^(-1)(1+x))]=1`

Simplify:

`[1/x+1/(x+1)]/[1-1/x*1/(x+1)]=1`

`(x+1+x)/[x(x+1)]= 1-1/(x^2+x)`

`(2x+1)/(x^2+x)=(x^2+x-1)/(x^2+x)`

`x^2-x-2=0`

`(x+1)(x-2)=0`

`x=-1, 2`