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Question Number 139609 by bemath last updated on 29/Apr/21

$$\mathrm{If}\tan 14°=\mathrm{x}\mathrm{then}\tan 18°=?$$

Answered by qaz last updated on 29/Apr/21

$$18°=\frac{18\times 14°}{14}=\frac{9}{7}{\tan }^{-1}x$$$$\Rightarrow \tan 18°=\tan \left(\frac{9}{7}{\tan }^{-1}x\right)$$

Answered by mr W last updated on 29/Apr/21

$$recall\tan 3\alpha =\frac{\tan \alpha (3-{\tan }^2\alpha )}{1-3{\tan }^2\alpha }$$$$$$$$\tan 42°=\tan \left(3\times 14°\right)=\frac{x(3-x^2)}{1-3x^2}$$$$\tan 18°=\tan (60°-42°)=\frac{\tan 60°-\tan 42°}{1+\tan 60°\tan 42°}$$$$=\frac{\sqrt{3}-\frac{x(3-x^2)}{1-3x^2}}{1+\sqrt{3}\times \frac{x(3-x^2)}{1-3x^2}}=\frac{\sqrt{3}-3x-3\sqrt{3}{x}^2+x^3}{1+3\sqrt{3}x-3x^2-\sqrt{3}{x}^3}$$

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