Question 126 of 155HardMCQ
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 :
Solutions · 1
Official solution
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 = - 432 Answer the question to unlock