Answer 1

`"Given ",tan^-1((x-1)/(x-2))+cot^-1((x+2)/(x+1))=pi/4`

`tan^-1((x-1)/(x-2))+tan^-1((x+1)/(x+2))=pi/4;`

`tan^-1((x-1)/(x-2))=pi/4-tan^-1((x+1)/(x+2))`

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

`=tan^-1[(1-(x+1)/(x+2))/(1+(x+1)/(x+2))]  ........[because tan^-1x-tan^-1y=tan^-1((x-y)/(1+xy))]`

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

`tan^-1((x-1)/(x-2))=tan^-1(1/(2x+3))`

`(x-1)/(x-2)=1/(2x+3)`

`(x-1)(2x+3)=x-2`

`2x^2-1=0`

`2x^2=1`

`x^2=1/2`

`x=+-1/sqrt2`