Verified Solution • 2025–26 Session moderate

QJEEM21Ma4188: JEE Main 2021 Mathematics - Class 11 JEEM21Ma4188 Solution

The value of cot⁡π24\cot {\pi \over {24}} is :

Multiple Choice Options

  • A 2+3+2−6\sqrt 2 + \sqrt 3 + 2 - \sqrt 6
  • B 2+3+2+6\sqrt 2 + \sqrt 3 + 2 + \sqrt 6
  • C 2−3−2+6\sqrt 2 - \sqrt 3 - 2 + \sqrt 6
  • D 32−3−63\sqrt 2 - \sqrt 3 - \sqrt 6

Step-by-Step Verified Solution

cot⁡θ=1+cos⁡2θsin⁡2θ=1+(3+122)(3−122)\cot \theta = {{1 + \cos 2\theta } \over {\sin 2\theta }} = {{1 + \left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right)} \over {\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right)}}

θ=π24\theta = {\pi \over {24}}

⇒cot⁡(π24)=1+(3+122)(3−122)\Rightarrow \cot \left( {{\pi \over {24}}} \right) = {{1 + \left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right)} \over {\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right)}}

=(22+3+1)(3−1)×(3+1)(3+1)= {{\left( {2\sqrt 2 + \sqrt 3 + 1} \right)} \over {\left( {\sqrt 3 - 1} \right)}} \times {{\left( {\sqrt 3 + 1} \right)} \over {\left( {\sqrt 3 + 1} \right)}}

=26+22+3+3+3+12= {{2\sqrt 6 + 2\sqrt 2 + 3 + \sqrt 3 + \sqrt 3 + 1} \over 2}

=6+2+3+2= \sqrt 6 + \sqrt 2 + \sqrt 3 + 2

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