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I have to find value of $f_{n} \theta = \tan \frac{\theta}{2} (1+\sec \theta)(1+\sec 2\theta)(1+\sec 4\theta)\dotsm(1+\sec 2^n\theta)$.

I tried to use $1 + \cos\theta = 2 \cos^2 \frac{\theta}{2}$, there were some cancellations but in end I got geometric progression of angles whose cosine were in product form.

How to deal with this?

Thanks

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    See also : http://math.stackexchange.com/questions/524742/trigonometry-simplify-and-find-the-value-of-tan-theta1-sec-frac-theta2?rq=1 – 2017-02-13

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We have $$f_0 \theta = \tan \frac {\theta}{2}(1+ \sec \theta) = \frac {\sin \theta/2}{\cos \theta/2} \frac {2\cos^2 \theta/2}{\cos \theta} = \frac {2\sin \theta/2 \cos \theta/2}{\cos \theta} = \tan \theta $$ Similarly, $$f_1 \theta = \tan \theta \frac {2\cos^2 \theta}{\cos 2\theta} = \tan 2\theta $$ $$f_2 \theta = \tan 2\theta \frac {2\cos^2 2\theta}{\cos 4\theta} = \tan 4\theta $$ $$\vdots $$

We can thus easily observe that $$f_n \theta = \tan 2^n \theta $$

Hope it helps.

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    How did you write $f_0 \theta$ – 2017-02-13
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    @Gathdi From the function definition $$f_n \theta = \tan \theta/2 \prod_{i=0}^{n} (1+ \sec 2^i \theta) $$ See what if $n=0$ and simplify. – 2017-02-13
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    but how can we generalise . maybe it changes at some stage. how can we be sure? – 2017-02-13
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    @Gathdi Just check for few values of $f_n \theta $ and confirm the result. – 2017-02-13