Please help me solve this question.(63. Question)

[Dexter1331]

 

[skeeter]

[MATH]100\theta = \dfrac{\pi}{2} \implies \theta = \dfrac{\pi}{200}[/MATH]
[MATH]\cot\left(\dfrac{n\pi}{200}\right) = \tan\left(\dfrac{100\pi}{200} - \dfrac{n\pi}{200}\right)[/MATH]
[MATH]\cot\left(\dfrac{\pi}{200} \right) \cdot \cot\left(\dfrac{2\pi}{200} \right) \cdot \cot\left(\dfrac{3\pi}{200} \right) \cdot \, ... \, \cdot \cot\left(\dfrac{50\pi}{200} \right) \cdot \tan\left(\dfrac{49\pi}{200} \right) \cdot \tan\left(\dfrac{48\pi}{200} \right) \cdot \, ... \, \cdot \tan\left(\dfrac{\pi}{200} \right) = ?[/MATH]

 

[Steven G]

[MATH]100\theta = \dfrac{\pi}{2} \implies \theta = \dfrac{\pi}{200}[/MATH]
[MATH]\cot\left(\dfrac{n\pi}{200}\right) = \tan\left(\dfrac{100\pi}{200} - \dfrac{n\pi}{200}\right)[/MATH]
[MATH]\cot\left(\dfrac{\pi}{200} \right) \cdot \cot\left(\dfrac{2\pi}{200} \right) \cdot \cot\left(\dfrac{3\pi}{200} \right) \cdot \, ... \, \cdot \cot\left(\dfrac{50\pi}{200} \right) \cdot \tan\left(\dfrac{49\pi}{200} \right) \cdot \tan\left(\dfrac{48\pi}{200} \right) \cdot \, ... \, \cdot \tan\left(\dfrac{\pi}{200} \right) = ?[/MATH]

I saw this trick before. I hope that the OP understands it because it really is an elegant way of seeing this problem.