cot 75° =___________________________ A. 2+\(\sqrt{3}\) B. 2-\(\sqrt{3}\)

cot-75deg-a-2-sqrt-3-b-2-sqrt-3

cot 75° =___________________________

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

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

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

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

This question was addressed to me in examination.

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

Correct answer is B. 2-\(\sqrt{3}\)

For 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 75° = (\(\sqrt{3}\) – 1)/(\(\sqrt{3}\) + 1) = 2-\(\sqrt{3}\).