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JEE Main 2021
Trigonometry
Trigonometric Ratio and Identites
Medium

Question

The value of cotπ24\cot {\pi \over {24}} is :

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Solution

cotθ=1+cos2θsin2θ=1+(3+122)(3122)\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)(3122) \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)(31)×(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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