The sum of possible values of x for

tan^$-$1(x + 1) + cot^$-$1$\left( {{1 \over {x - 1}}} \right)$ = tan^$-$1$\left( {{8 \over {31}}} \right)$ is :

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