prove that tan ((3π)/7) −4sin (π/7) = (√7).


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Question Number 69081 by myintkhaing1121960@gmail.com last updated on 19/Sep/19

$p r o v e t h a t$ $t a n \frac{3 π}{7} - 4 s i n \frac{π}{7} = \sqrt{7} .$
	Answered by Tanmay chaudhury last updated on 19/Sep/19

	$7 a = π$ $t a n 7 a = t a n π = - 0$ $\frac{s_{1} - s_{3} + s_{5} - s_{7}}{1 - s_{2} + s_{4} - s_{6}} = 0$ $7 c_{1} t a n a - 7 c_{3} {t a n}^{3} a + 7 c_{5} {t a n}^{5} a - 7 c_{7} {t a n}^{7} a = 0$ $7 t a n a - \frac{7 \times 6 \times 5}{3 \times 2} {t a n}^{3} a + \frac{7 \times 6}{2} {t a n}^{5} a - {t a n}^{7} a = 0$ $7 t a n a - 35 {t a n}^{3} a + 21 {t a n}^{5} a - {t a n}^{7} a = 0$ $t a n a = t a n \frac{π}{7} \neq 0$ $7 - 35 {t a n}^{2} a + 21 {t a n}^{4} a - {t a n}^{6} a = 0$ ${t a n}^{6} a - 21 {t a n}^{4} a + 35 {t a n}^{2} a - 7 = 0$ $f r o m a b o v e e q n w e h a v e t o f i n d v a l u e o f t a n a$ $t a n 3 a = \frac{3 t a n a - {t a n}^{3} a}{1 - 3 {t a n}^{2} a}$ $s i n a$ $= \sqrt{1 - \frac{1}{{s e c}^{2} a}}$ $= \sqrt{\frac{{t a n}^{2} a}{1 + {t a n}^{2} a}}$ $t a n 3 a - 4 s i n a$ $= \frac{3 t a n a - {t a n}^{3} a}{1 - 3 {t a n}^{2} a} - \frac{4 t a n a}{\sqrt{1 + {t a n}^{2} a}}$ $w a i t . . .$
	Answered by mind is power last updated on 19/Sep/19

	$w e h a v e$ $\underset{i = 1}{\prod^{m}} t a n (\frac{i π}{2 m + 1}) = \sqrt{2 m + 1}$ $w e t r y t o o a p p l i e t h i s g o r m = 3$ $\Rightarrow \tan (\frac{π}{7}) \tan (\frac{2 π}{7}) \tan (\frac{3 π}{7}) = \sqrt{7}$ $w e f i n i s h i f a n d o n l y i f$ $\Rightarrow \tan (\frac{π}{7}) \tan (\frac{2 π}{7}) \tan (\frac{3 π}{7}) = \tan (\frac{3 π}{7}) - 4 \sin (\frac{π}{7})$ $l e t a = \frac{π}{7}$ $w e w a n t t o o s h o w$ $\tan (a) \tan (2 a) \tan (3 a) = \tan (3 a) - 4 \sin (a)$ $\Leftrightarrow \frac{\sin (a) \sin (2 a) \sin (3 a)}{\cos (a) \cos (2 a) \cos (3 a)} = \frac{\sin (3 a) - 4 \sin (a) \cos (3 a)}{\cos (3 a)}$ $\Leftrightarrow \frac{\sin (a) . 2 \sin (a) \cos (a) \sin (3 a)}{\cos (a) \cos (2 a) \cos (3 a)} = \frac{\sin (3 a) - 2 . 2 \sin (a) c o s (3 a)}{\cos (3 a)}$ $\Leftrightarrow \frac{2 \sin^{2} (a) \sin (3 a)}{\cos (2 a)} = \sin (3 a) - 2 . 2 \cos (3 a) \sin (a)$ $\sin (x) \cos (y) = \frac{s i n (x + y) + s i n (x - y)}{2} \Rightarrow 2 \sin (a) \cos (3 a) = s i n (4 a) - s i n (2 a)$ $2 {s i n}^{2} (a) = 1 - c o s (2 a)$ $\Rightarrow$ $\frac{2 \sin^{2} (a) \sin (3 a)}{\cos (2 a)} = \sin (3 a) - 2 . 2 \cos (3 a) \sin (a)$ $\Leftrightarrow \frac{s i n (3 a) (1 - c o s (2 a))}{c o s (2 a)} = s i n (3 a) - 2 s i n (4 a) + 2 s i n (2 a)$ $\Leftrightarrow s i n (3 a) - s i n (3 a) c o s (2 a) = s i n (3 a) c o s (2 a) - 2 s i n (4 a) c o s (2 a) + 2 s i n (2 a) c o s (2 a)$ $\Leftrightarrow s i n (3 a) - \frac{s i n (5 a) + s i n (a)}{2} = \frac{s i n (5 a) + s i n (a)}{2} - s i n (6 a) - s i n (2 a) + s i n (4 a)$ $s i n (4 a) = s i n (3 a) . . . . a = \frac{π}{7}$ $s i n (6 a) = s i n (a)$ $s i n (5 a) = s i n (2 a)$ $\Rightarrow s i n (3 a) - \frac{s i n (2 a) + s i n (a)}{2} = \frac{s i n (2 a) + s i n (a)}{2} - s i n (a) - s i n (2 a) + s i n (3 a)$ $\Rightarrow 0 = \frac{s i n (2 a) + s i n (a) + s i n (2 a) + s i n (a)}{2} - s i n (a) - s i n (2 a) - s i n (3 a) + s i n (3 a)$ $\Leftrightarrow 0 = 0$ $\Rightarrow t a n (\frac{3 π}{7}) - 4 s i n (\frac{π}{7}) = t a n (\frac{π}{7}) t a n (\frac{2 π}{7}) t a n (\frac{3 π}{7}) = \underset{i = 1}{\prod^{3}} \tan (\frac{i π}{7}) = \sqrt{7}$
	Commented by MJS last updated on 19/Sep/19

	$great!$
	Commented by mind is power last updated on 19/Sep/19

	$t h a n k y o u$
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