JEE Mains · Maths · STD 11 - 3. trignometrical ratios,functions and identities
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 \(\sqrt{2}+\sqrt{3}+2+\sqrt{6}\)
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, \(\theta=\frac{\pi}{24}\)…
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