tan $-$1 (x + 1) + cot $-$1 $\left( {{1 \over {x - 1}}} \right)$ = tan $-$1 $\left( {{8 \over {31}}} \right)$ $\Rightarrow$ tan $-$1 (x + 1) + tan $-$1 (x - 1) = tan $-$1 $\left( {{8 \over {31}}} \right)$ $\Rightarrow$ ${\tan ^{ - 1}}\left( {{{\left( {x + 1} \right) + \left( {x - 1} \right)} \over {1 - \left( {x + 1} \right)\left( {x - 1} \right)}}} \right)$ = tan $-$1 $\left( {{8 \over {31}}} \right)$ $\Rightarrow$ ${{(1 + x) + (x - 1)} \over {1 - (1 + x)(x - 1)}} = {8 \over {31}}$ $\Rightarrow {{2x} \over {2 - {x^2}}} = {8 \over {31}}$ $\Rightarrow 4{x^2} + 31x - 8 = 0$ $\Rightarrow x = - 8,{1 \over 4}$ but at $x = {1 \over 4}$ $LHS > {\pi \over 2}$ and $RHS < {\pi \over 2}$ So, only solution is x = $-$ 8 = -$$$${{{32} \over 4}}