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Question Number 157388 by naka3546 last updated on 22/Oct/21

$$Ifm\tan (\theta -30°)=n\tan (\theta +12°)$$$$\cos 2\theta =?$$

Commented by cortano last updated on 22/Oct/21

$$m\tan (\theta -30°)=n\tan (30°+\theta -18°)$$$$\Rightarrow m\left(\frac{3\tan \theta -\sqrt{3}}{3+\sqrt{3}\tan \theta }\right)=n\left(\frac{\tan (30°+\theta )-\tan 18°}{1+\tan (30°+\theta )\tan 18°}\right)$$$$\tan 18°=\frac{\sin 18°}{\cos 18°}=\frac{\frac{\sqrt{5}-1}{4}}{\sqrt{1-{\left(\frac{\sqrt{5}-1}{4}\right)}^2}}$$$$=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}$$

Commented by naka3546 last updated on 23/Oct/21

$$thankyou,sir.$$

Answered by mr W last updated on 23/Oct/21

$$\frac{m}{n}\tan (\theta -30)=\tan (\theta -30+18)=\frac{\tan (\theta -30)+\tan 18}{1-\tan (\theta -30)\tan 18}$$$$\mu =\frac{m}{n},t=\tan (\theta -30)$$$$\mu t=\frac{t+\tan 18}{1-t\tan 18}$$$$\mu \tan 18t^2-(\mu -1)t+\tan 18=0$$$$t=\frac{\mu -1\pm \sqrt{{(\mu -1)}^2-4\mu {\tan }^{2}18}}{2\mu \tan 18}$$$$t=\frac{\tan \theta -\frac{1}{\sqrt{3}}}{1+\frac{\tan \theta }{\sqrt{3}}}=\frac{\sqrt{3}\tan \theta -1}{\tan \theta +\sqrt{3}}$$$$\tan \theta =\frac{1+\sqrt{3}t}{\sqrt{3}-t}$$$$\cos 2\theta =\frac{2}{1+{\tan }^2\theta }-1$$

Commented by naka3546 last updated on 23/Oct/21

$$\mathrm{Thank}\mathrm{you},\mathrm{sir}.$$

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