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Question Number 75131 by mathocean1 last updated on 07/Dec/19

$$\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{to}\mathrm{show}\mathrm{that}$$$${\tan }^2\left(\frac{\pi }{8}\right)+2\tan \left(\frac{\pi }{8}\right)-1=0$$

Answered by mr W last updated on 07/Dec/19

$$\tan \frac{\pi }{4}=1$$$$\Rightarrow \tan 2\times \frac{\pi }{8}=1$$$$\Rightarrow \frac{2\tan \frac{\pi }{8}}{1-{\tan }^2\frac{\pi }{8}}=1$$$$\Rightarrow 2\tan \frac{\pi }{8}=1-{\tan }^2\frac{\pi }{8}$$$$\Rightarrow {\tan }^2\frac{\pi }{8}+2\tan \frac{\pi }{8}-1=0$$

Commented by mathocean1 last updated on 07/Dec/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}...\mathrm{can}\mathrm{you}\mathrm{help}\mathrm{me}\mathrm{to}$$$$\mathrm{deduct}\tan \frac{\pi }{8}?$$$$$$$$$$$$$$

Commented by peter frank last updated on 07/Dec/19

$$\tan 2\theta =\frac{2\tan \theta }{1-\tan {}^2\theta }$$$$but\theta =\frac{\pi }{8}$$$$\tan 2\frac{\pi }{8}=\frac{2\tan \frac{\pi }{8}}{1-\tan {}^2\frac{\pi }{8}}$$$$\tan \frac{\pi }{4}=\frac{2\tan \frac{\pi }{8}}{1-\tan {}^2\frac{\pi }{8}}$$$$$$$$$$

Commented by MJS last updated on 07/Dec/19

$$\mathrm{right}\mathrm{idea},\mathrm{just}\mathrm{put}\theta =\frac{\alpha }{2}$$$$\tan \alpha =\frac{2\tan \frac{\alpha }{2}}{1-{\tan }^2\frac{\alpha }{2}}$$$$\mathrm{now}\mathrm{solve}\mathrm{for}\tan \frac{\alpha }{2}$$$$\tan \frac{\alpha }{2}=-\frac{1\pm \sqrt{1+{\tan }^2\alpha }}{\tan \alpha }$$$$\mathrm{now}\alpha =\frac{\pi }{4}\Rightarrow \tan \alpha =1$$$$\Rightarrow \tan \frac{\pi }{8}=-1+\sqrt{2}$$$$\mathrm{knowing}\tan \frac{\pi }{8}>0\mathrm{we}\mathrm{cannot}\mathrm{use}\mathrm{the}$$$$\mathrm{second}\mathrm{solution}-1-\sqrt{2}$$

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