The value of \cot \frac{\\pi}{24} is: [25 Jul 2021 Shift 2]

Correct Answer is : √2+√3+2+√6

\\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}

JEE Main 25 July 2021 Shift 2 Solved Paper

Section: Mathematics

Question : 66 of 90

Marks: +1, -0

The value of cot‌

is:

[25 Jul 2021 Shift 2]

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