JEE Main 2021TrigonometryTrigonometric Ratio and IdentitesMediumQuestionThe value of cotπ24\cot {\pi \over {24}}cot24π is :OptionsA2+3+2−6\sqrt 2 + \sqrt 3 + 2 - \sqrt 6 2+3+2−6B2+3+2+6\sqrt 2 + \sqrt 3 + 2 + \sqrt 6 2+3+2+6C2−3−2+6\sqrt 2 - \sqrt 3 - 2 + \sqrt 6 2−3−2+6D32−3−63\sqrt 2 - \sqrt 3 - \sqrt 6 32−3−6Check AnswerHide SolutionSolutioncotθ=1+cos2θsin2θ=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)}}cotθ=sin2θ1+cos2θ=(223−1)1+(223+1) θ=π24\theta = {\pi \over {24}}θ=24π ⇒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)}}⇒cot(24π)=(223−1)1+(223+1) =(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)}}=(3−1)(22+3+1)×(3+1)(3+1) =26+22+3+3+3+12 = {{2\sqrt 6 + 2\sqrt 2 + 3 + \sqrt 3 + \sqrt 3 + 1} \over 2}=226+22+3+3+3+1 =6+2+3+2 = \sqrt 6 + \sqrt 2 + \sqrt 3 + 2=6+2+3+2