We have,
`tan^(-1)(x+1)+tan^(-1)(x-1)=tan^(-1)8/31`
`rArr tan^(-1){((x+1)+(x-1))/(1-(x+1)(x-1))}=tan^(-1)8/31`
`rArr tan^(-1){((x+1)+(x-1))/(1-(x+1)(x-1))}=tan^(-1)8/31`
`rArr tan^(-1)((2x)/(2-x^(2))=tan^(-1)8/31`.
`rArr tan{(tan^(-1)((2x)/(2-x^(2))}=8/31 rArr (2x)/(2-x^(2))=8/31`
`rArr 8x^(2)+62x-16=0 rArr 4x^(2)+31x-8=0`
`rArr (4x-1)(x+8)=0 rArr x=1/4` or `x=-8`.
But, `x=-8` gives LHS `=tan^(-1)(-7)+tan^(-1)(-9)`,which is negative, while RHS is positive. So, `x=-8` is not possible.
Hence, `x=1/4`.
