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Question Number 173447 by ajfour last updated on 11/Jul/22

$$Find\tan (142.5°)without$$$$tables.$$

Commented by mr W last updated on 12/Jul/22

$$\mathit{w}\mathit{e}\mathit{l}\mathit{c}\mathit{o}\mathit{m}\mathit{e}\mathit{b}\mathit{a}\mathit{c}\mathit{k}\mathit{s}\mathit{i}\mathit{r}!$$$$$$$$lett=\tan 15°$$$$\tan 30°=\frac{2t}{1-t^2}=\frac{1}{\sqrt{3}}$$$$t^2+2\sqrt{3}t-1=0$$$$\Rightarrow t=2-\sqrt{3}$$$$$$$$lets=\tan 142.5°$$$$\frac{1}{\tan 15°}=\tan 75°=-\tan 285°=-\tan (2\times 142.5°)$$$$\frac{1}{2-\sqrt{3}}=-\frac{2s}{1-s^2}$$$$s^2-2(2-\sqrt{3})s-1=0$$$$\Rightarrow s=\tan 142.5°=2-\sqrt{3}-2\sqrt{2-\sqrt{3}}$$$$=(\sqrt{2}-\sqrt{3})(\sqrt{2}+1)$$

Commented by ajfour last updated on 11/Jul/22

$$\sqrt{2-\sqrt{3}}=p$$$$2-\sqrt{3}=p^2=\frac{1}{2+\sqrt{3}}$$$$p^2+\frac{1}{p^2}=4$$$${(p+\frac{1}{p})}^2=6$$$$p+\frac{1}{p}=\sqrt{6}$$$$p^2-\sqrt{6}p+1=0$$$$p=\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}$$$$hencerewritinganotherway$$$$s=2-\sqrt{3}-2p$$$$=2-\sqrt{3}-\sqrt{6}+\sqrt{2}$$$$thankyousir.$$$$$$$$$$

Commented by Tawa11 last updated on 11/Jul/22

$$\mathrm{Great}\mathrm{sirs}$$

Commented by Tawa11 last updated on 13/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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