cot 15° =______________ A. 2+\(\sqrt{3}\) B. 2-\(\sqrt{3}\) C.

cot-15deg-a-2-sqrt-3-b-2-sqrt-3-c

cot 15° =______________

A. 2+\(\sqrt{3}\)

B. 2-\(\sqrt{3}\)

C. 1+\(\sqrt{3}\)

D. \(\sqrt{3}\)-1

I have been asked this question during an interview for a job.

My enquiry is from Trigonometric Functions of Sum and Difference of Two Angles-1 topic in portion Trigonometric Functions of Mathematics – Class 11

Correct option is A. 2+\(\sqrt{3}\)

The explanation: We know, cot (x – y) = (cot x cot y +1)/cot y – cot x)

cot(45°-30°) = (cot 45° cot 30°+1)/(cot 45° – cot 30°)

cot 15° = (\(\sqrt{3}\) + 1)/(\(\sqrt{3}\) – 1) = 2+\(\sqrt{3}\).