Let Fn(θ)=tan⁡θ2(1+sec⁡θ)(1+sec⁡2θ)(1+sec⁡4θ)⋯1+sec⁡2nθ, then

∵Fn(θ)=tan⁡θ2(1+sec⁡θ)(1+sec⁡2θ)(1+sec⁡4θ)⋯1+sec⁡2nθ=tan⁡θ21+cos⁡θcos⁡θ1+cos⁡2θcos⁡2θ1+cos⁡4θcos⁡4θ⋯1+cos⁡2nθcos⁡2nθ=tan⁡θ2⋅2cos2⁡θ2⋅2cos2⁡θ2cos2⁡2θ⋯2cos2⁡2n−1θcos⁡θcos⁡2θcos⁡4θ⋯cos⁡2nθ=sin⁡θ2cos⁡θ2⋅2ncos2⁡θ2cos2⁡θcos2⁡2θ⋯cos2⁡2n−1θcos⁡θcos⁡2θcos⁡4θ⋯cos⁡2nθ=2nsin⁡θ2cos⁡θ2⋅cos⁡θcos⁡2θ⋯cos⁡2n−1θcos⁡2nθ=sin⁡2nθcos⁡2nθ=tan⁡2nθ (using formula) Fn(θ)=tan⁡2nθNow, F3π32=tan⁡23⋅π32=tan⁡π4=1 F4π64=tan⁡24⋅π64=tan⁡π4=1F5π128=tan⁡25⋅π128=tan⁡π4=1F6π256=tan⁡26⋅π256=tan⁡π4=1

Let Fn(θ)=tan⁡θ2(1+sec⁡θ)(1+sec⁡2θ)(1+sec⁡4θ)⋯1+sec⁡2nθ, then

∵Fn(θ)=tan⁡θ2(1+sec⁡θ)(1+sec⁡2θ)(1+sec⁡4θ)⋯1+sec⁡2nθ=tan⁡θ21+cos⁡θcos⁡θ1+cos⁡2θcos⁡2θ1+cos⁡4θcos⁡4θ⋯1+cos⁡2nθcos⁡2nθ=tan⁡θ2⋅2cos2⁡θ2⋅2cos2⁡θ2cos2⁡2θ⋯2cos2⁡2n−1θcos⁡θcos⁡2θcos⁡4θ⋯cos⁡2nθ=sin⁡θ2cos⁡θ2⋅2ncos2⁡θ2cos2⁡θcos2⁡2θ⋯cos2⁡2n−1θcos⁡θcos⁡2θcos⁡4θ⋯cos⁡2nθ=2nsin⁡θ2cos⁡θ2⋅cos⁡θcos⁡2θ⋯cos⁡2n−1θcos⁡2nθ=sin⁡2nθcos⁡2nθ=tan⁡2nθ (using formula) Fn(θ)=tan⁡2nθNow, F3π32=tan⁡23⋅π32=tan⁡π4=1 F4π64=tan⁡24⋅π64=tan⁡π4=1F5π128=tan⁡25⋅π128=tan⁡π4=1F6π256=tan⁡26⋅π256=tan⁡π4=1

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Let $F_{n} (θ) = \tan \frac{θ}{2} (1 + \sec θ) (1 + \sec 2 θ) (1 + \sec 4 θ) \dots (1 + \sec 2^{n} θ),$ then

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$F_{4} (\frac{π}{64}) = 1$

$F_{3} (\frac{π}{32}) = 1$

$F_{6} (\frac{π}{256}) = 1$

$F_{5} (\frac{π}{128}) = 1$

answer is A, B, C, D.

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Detailed Solution

$∵ F_{n} (θ) = \tan \frac{θ}{2} (1 + \sec θ) (1 + \sec 2 θ) (1 + \sec 4 θ) \dots (1 + \sec 2^{n} θ)$

$\begin{array}{l} = \tan \frac{θ}{2} (\frac{1 + \cos θ}{\cos θ}) (\frac{1 + \cos 2 θ}{\cos 2 θ}) (\frac{1 + \cos 4 θ}{\cos 4 θ}) \dots (\frac{1 + \cos 2^{n} θ}{\cos 2^{n} θ}) \\ = \tan \frac{θ}{2} \cdot \frac{2 \cos^{2} \frac{θ}{2} \cdot 2 \cos^{2} θ 2 \cos^{2} 2 θ \dots 2 \cos^{2} 2^{n - 1} θ}{\cos θ \cos 2 θ \cos 4 θ \dots \cos 2^{n} θ} \\ = \frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}} \cdot \frac{2^{n} \cos^{2} \frac{θ}{2} \cos^{2} θ \cos^{2} 2 θ \dots \cos^{2} 2^{n - 1} θ}{\cos θ \cos 2 θ \cos 4 θ \dots \cos 2^{n} θ} \end{array}$

$\begin{array}{l} = \frac{2^{n} \sin \frac{θ}{2} \cos \frac{θ}{2} \cdot \cos θ \cos 2 θ \dots \cos 2^{n - 1} θ}{\cos 2^{n} θ} \\ = \frac{\sin 2^{n} θ}{\cos 2^{n} θ} = \tan 2^{n} θ (u \sin g f o r m u l a) \end{array}$

$F_{n} (θ) = \tan 2^{n} θ$

Now, $F_{3} (\frac{π}{32}) = \tan (2^{3} \cdot \frac{π}{32}) = \tan \frac{π}{4} = 1$

$\begin{array}{l} F_{4} (\frac{π}{64}) = \tan (2^{4} \cdot \frac{π}{64}) = \tan \frac{π}{4} = 1 \\ F_{5} (\frac{π}{128}) = \tan (2^{5} \cdot \frac{π}{128}) = \tan \frac{π}{4} = 1 \\ F_{6} (\frac{π}{256}) = \tan (2^{6} \cdot \frac{π}{256}) = \tan \frac{π}{4} = 1 \end{array}$