Let f_ n ( )= 2 (1 + sec θ) (1 + sec 2θ) (1 + sec 4θ)............ (1 + sec 2 n θ), then
Mathematics · JEE Advanced · NTA Exams — Trigonometry
Let \(f_{\mathrm{n}}(\theta)=\tan \frac{\theta}{2}\)(1 + sec θ) (1 + sec 2θ)
(1 + sec 4θ)............ (1 + sec 2n θ), then
(1 + sec 4θ)............ (1 + sec 2n θ), then
- \(f_{2}\left(\frac{\pi}{16}\right)=1\)
- \(f_{3}\left(\frac{\pi}{32}\right)=1\)
- \(f_{4}\left(\frac{\pi}{64}\right)=1\)
- \(f_{5}\left(\frac{\pi}{128}\right)=1\)
Answer
(A) f_ 2 ( 16 )=1, (B) f_ 3 ( 32 )=1, (C) f_ 4 ( 64 )=1, (D) f_ 5 ( 128 )=1
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