prove it, tan (α+(π/3))+tan (α−(π/3))=((4sin 2α)/(1−4sin^2 α))


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Question Number 20647 by oyshi last updated on 30/Aug/17

$p r o v e i t,$ $\tan (α + \frac{π}{3}) + \tan (α - \frac{π}{3}) = \frac{4 \sin 2 α}{1 - 4 \sin^{2} α}$
	Answered by $@ty@m last updated on 30/Aug/17

	$L H S = \frac{t a n α + t a n \frac{π}{3}}{1 - t a n α . t a n \frac{π}{3}} + \frac{t a n α - t a n \frac{π}{3}}{1 + t a n α . t a n \frac{π}{3}}$ $= \frac{t a n α + \sqrt{3}}{1 - \sqrt{3} . t a n α} + \frac{t a n α - \sqrt{3}}{1 + \sqrt{3} . t a n α}$ $= \frac{(1 + \sqrt{3} . t a n α) (t a n α + \sqrt{3}) + (1 - \sqrt{3} . t a n α) (t a n α - \sqrt{3})}{(1 - \sqrt{3} . t a n α) (1 + \sqrt{3} . t a n α)}$ $= \frac{t a n α + \sqrt{3} {t a n}^{2} α + 3 t a n α + \sqrt{3} + t a n α - \sqrt{3} {t a n}^{2} α - \sqrt{3} + 3 t a n α}{1 - 3 {t a n}^{2} α}$ $= \frac{8 t a n α}{1 - 3 {t a n}^{2} α} = \frac{4 s i n 2 α}{{c o s}^{2} α - 3 {s i n}^{2} α}$ $(m u l t i p l y i n g N^{r} & D^{r} b y {c o s}^{2} α)$ $= \frac{4 s i n 2 α}{1 - {s i n}^{2} α - 3 {s i n}^{2} α}$ $= \frac{4 \sin 2 α}{1 - 4 \sin^{2} α}$ $= R H S$
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