ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 3. trignometrical ratios,functions and identities

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

  1. A \(\sqrt{2}-\sqrt{3}-2+\sqrt{6}\)
  2. B \(3 \sqrt{2}-\sqrt{3}-\sqrt{6}\)
  3. C \(\sqrt{2}-\sqrt{3}+2-\sqrt{6}\)
  4. D \(\sqrt{2}+\sqrt{3}+2+\sqrt{6}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\cot \theta=\frac{1+\cos 2 \theta}{\sin 2 \theta}=\{ \therefore 1+\cos 2 \theta=2 \cos ^{2} \theta \,\& \, \sin 2 \theta=2 \sin \theta \cos \theta\}\) put, \(\theta=\frac{\pi}{24}\)…
Same subject
Explore more questions on app