Myah Fuller

2022-02-27

$X=\left(1+{\mathrm{tan}1}^{\circ }\right)\left(1+{\mathrm{tan}2}^{\circ }\right)\left(1+{\mathrm{tan}3}^{\circ }\right)\dots \left(1+{\mathrm{tan}45}^{\circ }\right)$
$\mathrm{tan}\left(90-\theta \right)=\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta }$

Chettaf04

Let $\theta ={1}^{\circ }=\frac{\pi }{180}\text{rad}$. Then
$X=\prod _{k=1}^{45}\left(1+\mathrm{tan}\left(k\theta \right)\right)=\prod _{k=1}^{45}\frac{\mathrm{sin}\left(k\theta \right)+\mathrm{cos}\left(k\theta \right)}{\mathrm{cos}\left(k\theta \right)}$
$=\prod _{k=1}^{45}\sqrt{2}\frac{\mathrm{sin}\left(\left(k+45\right)\theta \right)}{\mathrm{cos}\left(k\theta \right)}$
Furthemore
$\prod _{k=1}^{45}\sqrt{2}\frac{\mathrm{sin}\left(\left(k+45\right)\theta \right)}{\mathrm{cos}\left(k\theta \right)}={2}^{\frac{45}{2}}\frac{\prod _{k=1}^{45}\mathrm{sin}\left(\left(k+45\right)\theta \right)}{\prod _{k=1}^{45}\mathrm{cos}\left(k\theta \right)}$
$={2}^{\frac{45}{2}}\frac{\prod _{n=0}^{44}\mathrm{sin}\left({90}^{\circ }-n\theta \right)}{\prod _{k=1}^{45}\mathrm{cos}\left(k\theta \right)}$
Now, using $\mathrm{sin}\left({90}^{\circ }-\alpha \right)=\mathrm{cos}\left(\alpha \right)$
${2}^{\frac{45}{2}}\frac{\prod _{n=0}^{44}\mathrm{sin}\left({90}^{\circ }-n\theta \right)}{\prod _{k=1}^{45}\mathrm{cos}\left(k\theta \right)}={2}^{\frac{45}{2}}\frac{\prod _{n=0}^{44}\mathrm{cos}\left(n\theta \right)}{\prod _{k=1}^{45}\mathrm{cos}\left(k\theta \right)}$
$={2}^{\frac{45}{2}}\frac{\mathrm{cos}\left(0\right)}{\mathrm{cos}\left({45}^{\circ }\right\}}\right\}={2}^{23}$

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