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Question Number 107672 by bemath last updated on 12/Aug/20

$$\mathrm{♠}\mathcal{B}e\mathcal{M}ath\mathrm{♠}$$$$\frac{4\sin \left(\frac{2\pi }{7}\right)+\sec \left(\frac{\pi }{14}\right)}{\cot \left(\frac{\pi }{7}\right)}?$$

Commented by Her_Majesty last updated on 12/Aug/20

$$2$$

Commented by ajfour last updated on 12/Aug/20

$$HerMJStywouldyoupleaseexplain$$$$thistoo.$$

Answered by 1549442205PVT last updated on 12/Aug/20

$$\mathrm{We}\mathrm{prove}\mathrm{that}\frac{4\sin \left(\frac{2\pi }{7}\right)+\sec \left(\frac{\pi }{14}\right)}{\cot \left(\frac{\pi }{7}\right)}=2(\ast )$$$$\iff \tan \left(\frac{\pi }{7}\right)[4\sin \left(\frac{2\pi }{7}\right)+\sec \left(\frac{\pi }{14}\right)]=2$$$$\iff 4\sin \frac{\pi }{7}\cos \frac{\pi }{14}\sin \frac{2\pi }{7}+\sin \frac{\pi }{7}=2\cos \frac{\pi }{7}\cos \frac{\pi }{14}$$$$\iff 2(\sin \frac{3\pi }{14}+\sin \frac{\pi }{14})\sin \frac{2\pi }{7}+\sin \frac{\pi }{7}$$$$=\cos \frac{3\pi }{14}+\cos \frac{\pi }{14}$$$$\iff \left(2\sin \frac{3\pi }{14}\sin \frac{2\pi }{7}\right)+\left(2\sin \frac{2\pi }{7}\sin \frac{\pi }{14}\right)\sin \frac{2\pi }{7}+\sin \frac{\pi }{7}-(\cos \frac{3\pi }{14}+\cos \frac{\pi }{14})=0$$$$\iff \cos \frac{\pi }{14}-\cos \frac{7\pi }{14}+\cos \frac{3\pi }{14}-\cos \frac{5\pi }{14}+\sin \frac{\pi }{7}+\sin \frac{2\pi }{7}-(\cos \frac{3\pi }{14}+\cos \frac{\pi }{14})=0$$$$\iff \sin \frac{\pi }{7}-\cos \frac{5\pi }{14}=0(\mathrm{since}\cos \frac{7\pi }{14}=\cos \frac{\pi }{2}=0)$$$$\mathrm{This}\mathrm{equlity}\mathrm{is}\mathrm{always}\mathrm{true}\mathrm{since}$$$$\sin \frac{\pi }{7}=\cos (\frac{\pi }{2}-\frac{\pi }{7})=\cos \frac{5\pi }{14}$$$$\mathrm{Thus},\mathrm{the}\mathrm{equality}(\ast )\mathrm{proved}$$

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