Verified Solution • 2025–26 Session moderate

QJEEM21Mad733: JEE Main 2021 Mathematics - Class 12 JEEM21Mad733 Solution: The sum of possible values of x for tan 1 (x + 1) + cot 1 = tan …

The sum of possible values of x for

tan −- 1(x + 1) + cot −- 1 (1x−1)\left( {{1 \over {x - 1}}} \right) = tan −- 1 (831)\left( {{8 \over {31}}} \right) is :

Multiple Choice Options

  • A −- 324{{{32} \over 4}}
  • B −- 334{{{33} \over 4}}
  • C −- 314{{{31} \over 4}}
  • D −- 304{{{30} \over 4}}

Step-by-Step Verified Solution

tan −- 1(x + 1) + cot −- 1 (1x−1)\left( {{1 \over {x - 1}}} \right) = tan −- 1 (831)\left( {{8 \over {31}}} \right)

⇒\Rightarrow tan −- 1(x + 1) + tan −- 1(x - 1) = tan −- 1 (831)\left( {{8 \over {31}}} \right)

⇒\Rightarrow tan⁡−1((x+1)+(x−1)1−(x+1)(x−1)){\tan ^{ - 1}}\left( {{{\left( {x + 1} \right) + \left( {x - 1} \right)} \over {1 - \left( {x + 1} \right)\left( {x - 1} \right)}}} \right) = tan −- 1 (831)\left( {{8 \over {31}}} \right)

⇒\Rightarrow (1+x)+(x−1)1−(1+x)(x−1)=831{{(1 + x) + (x - 1)} \over {1 - (1 + x)(x - 1)}} = {8 \over {31}}

⇒2x2−x2=831\Rightarrow {{2x} \over {2 - {x^2}}} = {8 \over {31}}

⇒4x2+31x−8=0\Rightarrow 4{x^2} + 31x - 8 = 0

⇒x=−8,14\Rightarrow x = - 8,{1 \over 4}

but at x=14x = {1 \over 4}

LHS > {\pi \over 2} and RHS < {\pi \over 2}

So, only solution is x = −- 8 = −- 324{{{32} \over 4}}

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