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Question Number 143384 by mathdanisur last updated on 13/Jun/21

$$proof:{tg}^2\left(36°\right)\cdot {tg}^2\left(72°\right)=5$$

Answered by Dwaipayan Shikari last updated on 13/Jun/21

$$tan36°=\frac{sin\left(36°\right)}{cos\left(36°\right)}=\frac{1-cos72°}{sin72°}=\frac{1-\frac{\sqrt{5}-1}{4}}{\frac{1}{4}\sqrt{10-2\sqrt{5}}}=\frac{5-\sqrt{5}}{\sqrt{10-2\sqrt{5}}}$$$$tan72°=\frac{cos18°}{sin18°}=\frac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}-1}$$$${tan}^{2}72°{tan}^{2}36°={\left(\frac{5-\sqrt{5}}{\sqrt{5}-1}\right)}^2=5$$

Commented by mathdanisur last updated on 14/Jun/21

$$Sirthankyou$$

Answered by Ar Brandon last updated on 13/Jun/21

$$\alpha ={\tan }^2\left(\frac{\pi }{5}\right){\tan }^2\left(\frac{2\pi }{5}\right)={\tan }^2\left(\frac{\pi }{5}\right){\cot }^2\left(\frac{\pi }{10}\right)$$$$={\left(\frac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}+1}\right)}^2{\left(\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1}\right)}^2=\frac{80}{16}=5$$

Commented by mathdanisur last updated on 14/Jun/21

$$thankyouSir$$

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