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
Best Answer
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.