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Question Number 143499 by mnjuly1970 last updated on 15/Jun/21

Commented by MJS_new last updated on 15/Jun/21

$$$$$$\tan (60°-\alpha )=\frac{\sqrt{3}-\tan \alpha }{1+\sqrt{3}\tan \alpha }$$$$\tan (60°+\alpha )=\frac{\sqrt{3}+\tan \alpha }{1-\sqrt{3}\tan \alpha }$$$$\tan 3\alpha =\frac{(3-{\tan }^2\alpha )\tan \alpha }{1-{3\tan }^2\alpha }$$$$\mathrm{now}\mathrm{simply}\mathrm{let}\tan \alpha =t$$

Answered by justtry last updated on 15/Jun/21

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