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Question Number 189482 by MATHEMATICSAM last updated on 17/Mar/23

$$\mathrm{If}\tan 11°=x\mathrm{then}\tan 1°=?$$

Answered by Frix last updated on 17/Mar/23

$$\tan 11°=\tan (9°+2°)=x$$$$=\frac{\tan 9°+\tan 2°}{1-\tan 9°\tan 2°}=x$$$$\tan 2°=\frac{x-\tan 9°}{1+x\tan 9°}=y$$$$\tan \left(2\times 1°\right)=\frac{2\tan 1°}{1-{\tan }^{2}1°}=y$$$$y>0\wedge \tan 1°>0\Rightarrow$$$$\tan 1°=\frac{\sqrt{y^2+1}-1}{y}=$$$$=\frac{\sqrt{(1+{\tan }^{2}9°)(x^2+1)}-(1+x\tan 9°)}{x-\tan 9°}$$$$\tan 18°=\frac{\sqrt{5(5-2\sqrt{5})}}{5}$$$$\tan 9°=\tan \frac{18°}{2}=\frac{\sqrt{1+{\tan }^{2}18°}-1}{\tan 18°}=...$$$$=\sqrt{2(3+\sqrt{5})}-\sqrt{5+2\sqrt{5}}$$$$$$$$\mathrm{Result}:$$$$\tan 1°=\frac{\sqrt{(1+{\tan }^{2}9°)(x^2+1)}-(1+x\tan 9°)}{x-\tan 9°}$$$$\mathrm{with}\tan 9°=\sqrt{2(3+\sqrt{5})}-\sqrt{5+2\sqrt{5}}$$

Answered by mr W last updated on 17/Mar/23

$$\tan 11°=x$$$$\tan 22°=\frac{2x}{1-x^2}$$$$\tan 44°=\frac{2\left(\frac{2x}{1-x^2}\right)}{1-{\left(\frac{2x}{1-x^2}\right)}^2}=\frac{4x(1-x^2)}{1-6x^2+x^4}$$$$\tan 1°=\tan (45°-44°)=\frac{1-\frac{4x(1-x^2)}{1-6x^2+x^4}}{1+\frac{4x(1-x^2)}{1-6x^2+x^4}}$$$$=\frac{1-4x-6x^2+4x^3+x^4}{1+4x-6x^2-4x^3+x^4}✓$$

Commented by Frix last updated on 17/Mar/23

$$\mathrm{Nice},\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{of}\mathrm{this}!$$

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