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Question Number 134417 by deleteduser12 last updated on 03/Mar/21

Answered by som(math1967) last updated on 03/Mar/21

$$\frac{2^{n-1}2sin\theta cos\theta .cos2\theta ...cos{2}^{n-1}\theta }{sin\theta }$$$$\frac{2^{n-2}.2sin2\theta cos2\theta .cos{2}^2\theta .....}{sin\theta }$$$$\frac{2^{n-3}2sin{2}^2\theta cos{2}^2\theta ...}{sin\theta }$$$$\frac{sin{2}^n\theta }{sin\theta }$$$$nowgiven\theta =\frac{\pi }{2^n+1}$$$$2^n\theta =\pi -\theta$$$$sin{2}^n\theta =sin(\pi -\theta )=sin\theta$$$$\therefore \frac{sin{2}^n\theta }{sin\theta }=\frac{sin\theta }{sin\theta }=1$$

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