JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 5)
The value of $$\cot {\pi \over {24}}$$ is :
$$\sqrt 2 + \sqrt 3 + 2 - \sqrt 6 $$
$$\sqrt 2 + \sqrt 3 + 2 + \sqrt 6 $$
$$\sqrt 2 - \sqrt 3 - 2 + \sqrt 6 $$
$$3\sqrt 2 - \sqrt 3 - \sqrt 6 $$
Explanation
$$\cot \theta = {{1 + \cos 2\theta } \over {\sin 2\theta }} = {{1 + \left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right)} \over {\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right)}}$$
$$\theta = {\pi \over {24}}$$
$$ \Rightarrow \cot \left( {{\pi \over {24}}} \right) = {{1 + \left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right)} \over {\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right)}}$$
$$ = {{\left( {2\sqrt 2 + \sqrt 3 + 1} \right)} \over {\left( {\sqrt 3 - 1} \right)}} \times {{\left( {\sqrt 3 + 1} \right)} \over {\left( {\sqrt 3 + 1} \right)}}$$
$$ = {{2\sqrt 6 + 2\sqrt 2 + 3 + \sqrt 3 + \sqrt 3 + 1} \over 2}$$
$$ = \sqrt 6 + \sqrt 2 + \sqrt 3 + 2$$
$$\theta = {\pi \over {24}}$$
$$ \Rightarrow \cot \left( {{\pi \over {24}}} \right) = {{1 + \left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right)} \over {\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right)}}$$
$$ = {{\left( {2\sqrt 2 + \sqrt 3 + 1} \right)} \over {\left( {\sqrt 3 - 1} \right)}} \times {{\left( {\sqrt 3 + 1} \right)} \over {\left( {\sqrt 3 + 1} \right)}}$$
$$ = {{2\sqrt 6 + 2\sqrt 2 + 3 + \sqrt 3 + \sqrt 3 + 1} \over 2}$$
$$ = \sqrt 6 + \sqrt 2 + \sqrt 3 + 2$$
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