The value of $\cot \frac{\pi}{24}$ is :

(A) $\sqrt{2}+\sqrt{3}+2+\sqrt{6}$

(B) $3 \sqrt{2}-\sqrt{3}-\sqrt{6}$

(C) $\sqrt{2}-\sqrt{3}-2+\sqrt{6}$

(D) $\sqrt{2}+\sqrt{3}+2-\sqrt{6}$

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Trigonometric Ratios and Identities

The value of

cot \frac{π}{24}

is :

Answer: (A)

Using

cot \frac{θ}{2} = \frac{1 + cos θ}{sin θ}

with

θ = \frac{π}{12}

, together with

cos \frac{π}{12} = \frac{6 + 2}{4}

and

sin \frac{π}{12} = \frac{6 - 2}{4}

cot \frac{π}{24} = \frac{1 + \frac{6 + 2}{4}}{\frac{6 - 2}{4}} = \frac{4 + 6 + 2}{6 - 2}

Multiplying numerator and denominator by

6 + 2

makes the denominator

4

and the numerator

8 + 42 + 43 + 46

. Dividing through by

4

gives

cot \frac{π}{24} = 2 + 3 + 2 + 6

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option