For a positive integer n, let fn(θ)=tan⁡θ2(1+sec⁡θ)(1+sec⁡2θ)…1+sec⁡2nθ then

∵tan⁡θ2(1+sec⁡θ)=tan⁡θ21+cos⁡θcos⁡θ=sin⁡θ2cos⁡θ2⋅2cos2⁡θ2cos⁡θ=2sin⁡θ2⋅cos⁡θ2cos⁡θ=sin⁡θcos⁡θ=tan⁡θ----iBy repeated use of Eq. (i), we havefn(θ)=tan⁡θ(1+sec⁡2θ)(1+sec⁡4θ)…1+sec⁡2nθ=tan⁡2θ(1+sec⁡4θ)…1+sec⁡2nθ=tan⁡4θ(1+sec⁡8θ)…1+sec⁡2nθ=….=tan⁡2nθ Now ,f2π16=tan⁡22π16=tan⁡π4=1f3π32=tan⁡23π32=tan⁡π4=1f4π64=tan⁡24π64=tan⁡π4=1and f5π128=tan⁡π4=1

For a positive integer n, let fn(θ)=tan⁡θ2(1+sec⁡θ)(1+sec⁡2θ)…1+sec⁡2nθ then

∵tan⁡θ2(1+sec⁡θ)=tan⁡θ21+cos⁡θcos⁡θ=sin⁡θ2cos⁡θ2⋅2cos2⁡θ2cos⁡θ=2sin⁡θ2⋅cos⁡θ2cos⁡θ=sin⁡θcos⁡θ=tan⁡θ----iBy repeated use of Eq. (i), we havefn(θ)=tan⁡θ(1+sec⁡2θ)(1+sec⁡4θ)…1+sec⁡2nθ=tan⁡2θ(1+sec⁡4θ)…1+sec⁡2nθ=tan⁡4θ(1+sec⁡8θ)…1+sec⁡2nθ=….=tan⁡2nθ Now ,f2π16=tan⁡22π16=tan⁡π4=1f3π32=tan⁡23π32=tan⁡π4=1f4π64=tan⁡24π64=tan⁡π4=1and f5π128=tan⁡π4=1

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For a positive integer n, let $f_{n} (θ) = \frac{\tan θ}{2} (1 + \sec θ) (1 + \sec 2 θ) \dots (1 + \sec 2^{n} θ)$ then

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$f_{5} (\frac{π}{128}) = 1$

$f_{2} (\frac{π}{16}) = 0$

$f_{4} (\frac{π}{64}) = - 1$

$f_{3} (\frac{π}{32}) = - 1$

answer is D.

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Detailed Solution

$\begin{array}{l} ∵ (\tan \frac{θ}{2}) (1 + \sec θ) = \tan (\frac{θ}{2}) (\frac{1 + \cos θ}{\cos θ}) \\ = \frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}} \cdot \frac{2 \cos^{2} \frac{θ}{2}}{\cos θ} = \frac{2 \sin \frac{θ}{2} \cdot \cos \frac{θ}{2}}{\cos θ} \\ = \frac{\sin θ}{\cos θ} = \tan θ - - - - (i) \end{array}$
By repeated use of Eq. (i), we have
$\begin{matrix} f_{n} (θ) = \tan θ (1 + \sec 2 θ) (1 + \sec 4 θ) \dots (1 + \sec 2^{n} θ) \\ = \tan 2 θ (1 + \sec 4 θ) \dots (1 + \sec 2^{n} θ) \end{matrix}$
$\begin{array}{l} = \tan 4 θ (1 + \sec 8 θ) \dots (1 + \sec 2^{n} θ) = \dots . = \tan 2^{n} θ \\ Now, f_{2} (\frac{π}{16}) = \tan (2^{2} \frac{π}{16}) = \tan \frac{π}{4} = 1 \\ f_{3} (\frac{π}{32}) = \tan (2^{3} \frac{π}{32}) = \tan \frac{π}{4} = 1 \\ f_{4} (\frac{π}{64}) = \tan (2^{4} \frac{π}{64}) = \tan \frac{π}{4} = 1 \\ and f_{5} (\frac{π}{128}) = \tan \frac{π}{4} = 1 \end{array}$