The value of 2+ 2+ 2+2 8 . where 0< < 8 is equal to
Mathematics · JEE Main · NTA Exams — Trigonometry
The value of \(\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}} .\) where \(0<\theta<\frac{\pi}{8}\) is equal to
- \(2 \cos \theta\)
- \(\cos \theta\)
- \(2 \sin \theta\)
- \(-2 \sin \theta\)
Answer
(A) 2
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