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Question Number 139555 by snipers237 last updated on 28/Apr/21

$$\underset{i=0}{\prod ^5}(1-cotan(20+i))\stackrel{?}{=}8$$

Answered by mr W last updated on 28/Apr/21

$$(1-\frac{1}{\tan 20°})(1-\frac{1}{\tan 25°})$$$$=1-\frac{1}{\tan 20°}-\frac{1}{\tan 25°}+\frac{1}{\tan 20°\tan 25°}$$$$=1-\frac{\tan 20°+\tan 25°}{\tan 20°\tan 25°}+\frac{1}{\tan 20°\tan 25°}$$$$=1-\frac{\tan 20°+\tan 25°}{\tan 20°\tan 25°}\times \frac{1-\tan 20°\tan 25°}{1-\tan 20°\tan 25°}+\frac{1}{\tan 20°\tan 25°}$$$$=1-\frac{\tan 45°(1-\tan 20°\tan 25°)}{\tan 20°\tan 25°}+\frac{1}{\tan 20°\tan 25°}$$$$=1-\frac{1-\tan 20°\tan 25°}{\tan 20°\tan 25°}+\frac{1}{\tan 20°\tan 25°}$$$$=1-\frac{1}{\tan 20°\tan 25°}+1+\frac{1}{\tan 20°\tan 25°}$$$$=2$$$$similarly$$$$(1-\frac{1}{\tan 21°})(1-\frac{1}{\tan 24°})=2$$$$(1-\frac{1}{\tan 22°})(1-\frac{1}{\tan 23°})=2$$$$\Rightarrow \underset{i=0}{\prod ^5}(1-\cot (20+i))$$$$=(1-\frac{1}{\tan 20°})(1-\frac{1}{\tan 21°})(1-\frac{1}{\tan 22°})(1-\frac{1}{\tan 23°})(1-\frac{1}{\tan 24°})(1-\frac{1}{\tan 25°})$$$$=2\times 2\times 2$$$$=8$$

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