if fn(θ)=tanθ2(1+sec⁡θ)(1+sec⁡2θ)…1+sec⁡2nθ,then

# if ${f}_{n}\left(\theta \right)=\mathrm{tan}\frac{\theta }{2}\left(1+\mathrm{sec}\theta \right)\left(1+\mathrm{sec}2\theta \right)\dots \left(1+\mathrm{sec}{2}^{n}\theta \right),$then

1. A

${f}_{2}\left(\frac{\pi }{16}\right)=1$

2. B

${f}_{3}\left(\frac{\pi }{32}\right)=1$

3. C

${f}_{4}\left(\frac{\pi }{64}\right)=1$

4. D

${f}_{5}\left(\frac{\pi }{128}\right)=1$

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### Solution:

${f}_{0}\left(\theta \right)=t\mathrm{an}\frac{\theta }{2}\left(1+\mathrm{sec}\theta \right)$

$=\frac{\mathrm{sin}\frac{\theta }{2}}{\mathrm{cos}\frac{\theta }{2}}\left(1+\frac{1}{\mathrm{cos}\theta }\right)=\frac{\mathrm{sin}\frac{\theta }{2}}{\mathrm{cos}\frac{\theta }{2}}\frac{\left(\mathrm{cos}\theta +1\right)}{\mathrm{cos}\theta }$

$\begin{array}{l}=\frac{2\mathrm{sin}\frac{\theta }{2}\mathrm{cos}\frac{\theta }{2}}{\mathrm{cos}\theta }=\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=\mathrm{tan}\theta \\ {f}_{1}\left(\theta \right)=\mathrm{tan}\theta \left(1+\mathrm{sec}2\theta \right)=\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }\frac{\left(\mathrm{cos}2\theta +1\right)}{\mathrm{cos}2\theta }=\mathrm{tan}2\theta \\ \mathrm{Likewise}{f}_{n}\left(\theta \right)=\mathrm{tan}\left({2}^{n}\theta \right)\\ {f}_{2}\left(\frac{\pi }{16}\right)=\mathrm{tan}\left({2}^{2}\cdot \frac{\pi }{16}\right)=1,{f}_{3}\left(\frac{\pi }{32}\right)=\mathrm{tan}\left({2}^{3}\cdot \frac{\pi }{32}\right)=1\\ {f}_{4}\left(\frac{\pi }{64}\right)=\mathrm{tan}\left({2}^{4}\cdot \frac{\pi }{64}\right)=1,{f}_{5}\left(\frac{\pi }{128}\right)=\mathrm{tan}\left({2}^{5}\cdot \frac{\pi }{128}\right)=1\end{array}$

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