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$$X=(1 + \tan 1^{\circ})(1 + \tan 2^{\circ})(1 + \tan 3^{\circ})\ldots(1 + \tan {45}^{\circ})$$

$$\tan(90-\theta)=\cot\theta=\frac{1}{\tan\theta}$$

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marked as duplicate by David Mitra, Sasha, Argon, Henry T. Horton, Asaf Karagila Feb 15 '13 at 1:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    

1 Answer 1

Let $\theta = 1^\circ = \frac{\pi}{180} \operatorname{rad}$. Then$$ X = \prod_{k=1}^{45} \left(1 + \tan(k \theta)\right) = \prod_{k=1}^{45} \frac{\sin(k \theta) + \cos(k \theta)}{\cos(k \theta)} = \prod_{k=1}^{45} \sqrt{2} \frac{\sin((k+45) \theta)}{\cos(k \theta)} $$ Furthermore $$ \prod_{k=1}^{45} \sqrt{2} \frac{\sin((k+45) \theta)}{\cos(k \theta)} = 2^{45/2} \frac{\prod_{k=1}^{45} \sin((k+45) \theta)}{\prod_{k=1}^{45} \cos(k \theta)} \stackrel{45+k = 90-n}{=} 2^{45/2} \frac{\prod_{n=0}^{44} \sin(90^\circ-n \theta)}{\prod_{k=1}^{45} \cos(k \theta)} $$ Now, using $\sin(90^\circ - \alpha) = \cos(\alpha)$: $$ 2^{45/2} \frac{\prod_{n=0}^{44} \sin(90^\circ-n \theta)}{\prod_{k=1}^{45} \cos(k \theta)} = 2^{45/2} \frac{\prod_{n=0}^{44} \cos(n \theta)}{\prod_{k=1}^{45} \cos(k \theta)} = 2^{45/2} \frac{\cos( 0 )}{\cos(45^\circ)} = 2^{23} $$

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Great. Note this is essentially the same proof as the first one in the duplicate. –  1015 Feb 15 '13 at 1:30
    
@julien Thanks for the heads up. I have not looked. This question should be closed as a duplicate, no doubt. –  Sasha Feb 15 '13 at 1:32

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