prove that (tan 4a+tan 2a)(1−tan^2 3a tan^2 a)=2tan 3a sec^2 a


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Question Number 39312 by kunal1234523 last updated on 05/Jul/18

$p r o v e t h a t$ $(t a n 4 a + t a n 2 a) (1 - {t a n}^{2} 3 a {t a n}^{2} a) = 2 t a n 3 a {s e c}^{2} a$
	Answered by kunal1234523 last updated on 05/Jul/18

	Commented by kunal1234523 last updated on 05/Jul/18

	$a n o t h e r w a y p l e a s e$
	Commented by abdo mathsup 649 cc last updated on 08/Jul/18

	$a n o t h e r w a y i s g o i n g t o h o s p i t a l o f c r a z y . . .$
	Answered by MJS last updated on 05/Jul/18

	$let me write c_{n} / s_{n} / t_{n} for \cos n x / \sin n x / \tan n x$ $(t_{4} + t_{2}) (1 - t_{3}^{2} t^{2}) = 2 \frac{t_{3}}{c^{2}}$ $t_{2} + t_{4} - t^{2} t_{3}^{2} t_{4} - t^{2} t_{2} t_{3}^{2} = 2 \frac{t_{3}}{c^{2}}$ $\frac{s_{2}}{c_{2}} + \frac{s_{4}}{c_{4}} - \frac{s^{2} s_{3}^{2} s_{4}}{c^{2} c_{3}^{2} c_{4}} - \frac{s^{2} s_{2} s_{3}^{2}}{c^{2} c_{2} c_{3}^{2}} = 2 \frac{s_{3}}{c^{2} c_{3}}$ $s_{2} c^{2} c_{3}^{2} c_{4} + s_{4} c^{2} c_{2} c_{3}^{2} - s^{2} s_{3}^{2} s_{4} c_{2} - s^{2} s_{2} s_{3}^{2} c_{4} = 2 s_{3} c_{2} c_{3} c_{4}$ $(c^{2} c_{3}^{2} - s^{2} s_{3}^{2}) (s_{2} c_{4} + s_{4} c_{2}) = 2 s_{3} c_{2} c_{3} c_{4}$ $(c^{2} c_{3}^{2} - (1 - c^{2}) (1 - c_{3}^{2})) (s_{2} c_{4} + s_{4} c_{2}) = 2 s_{3} c_{2} c_{3} c_{4}$ $(c^{2} + c_{3}^{2} - 1) (s_{2} c_{4} + s_{4} c_{2}) = 2 s_{3} c_{2} c_{3} c_{4}$ $now use trigonometric formulas until you$ $reach this . . .$ $(\frac{1}{2} (c_{2} + c_{6})) (s_{6}) = \frac{1}{4} (s_{4} + s_{8} + s_{12})$ $. . . and again :$ $\frac{1}{4} (s_{4} + s_{8} + s_{12}) = \frac{1}{4} (s_{4} + s_{8} + s_{12})$
	Answered by MJS last updated on 05/Jul/18

	$t = \arctan α$ $\tan α \to t$ $\tan 2 α \to \frac{2 t}{(1 - t) (1 + t)}$ $\tan 3 α \to \frac{t (3 - t^{2})}{1 - 3 t^{2}}$ $\tan 4 α \to \frac{4 t (1 - t) (1 + t)}{(1 + 2 t - t^{2}) (1 - 2 t - t^{2})}$ $\sec^{2} α \to 1 + t^{2}$ $left side$ $(\frac{2 t}{(1 - t) (1 + t)} + \frac{4 t (1 - t) (1 + t)}{(1 + 2 t - t^{2}) (1 - 2 t - t^{2})}) =$ $= \frac{2 t (1 - 3 t^{2}) (3 - t^{2})}{(1 + 2 t - t^{2}) (1 - 2 t - t^{2}) (1 - t) (1 + t)} = A$ $(1 - {(\frac{t (3 - t^{2})}{1 - 3 t^{2}})}^{2} t^{2}) = \frac{(1 - 6 t^{2} + t^{4}) (1 + t^{2}) (1 - t) (1 + t)}{{(1 - 3 t^{2})}^{2}} = B$ $A \times B = \frac{2 t (3 - t^{2}) (1 + t^{2})}{1 - 3 t^{2}}$ $right side$ $2 \frac{t (3 - t^{2})}{1 - 3 t^{2}} (1 + t^{2})$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jul/18

	$(\frac{s i n 4 α}{c o s 4 α} + \frac{s i n 2 α}{c o s 2 α}) (\frac{{c o s}^{2} 3 α {c o s}^{2} α - {s i n}^{2} 3 α {s i n}^{2} α}{{c o s}^{2} 3 α {c o s}^{2} α})$ $(\frac{s i n 6 α}{c o s 4 α c o s 2 α}) (\frac{\frac{1 + c 6 α}{2} . \frac{1 + c 2 α}{2} - \frac{1 - c 6 α}{2} . \frac{1 - c 2 α}{2}}{c^{2} 3 α . c^{2} α})$ $\frac{2 s 3 α c 3 α}{c 4 α c 2 α} . \frac{1}{4} . \frac{2 c 6 α + 2 c 2 α}{{c o s}^{2} 3 α {c o s}^{2} α}$ $\frac{s i n 3 α c o s 3 α}{c o s 4 α c o s 2 α} . \frac{c o s 6 α + c o s 2 α}{{c o s}^{2} 3 α {c o s}^{2} α}$ $\frac{s i n 3 α c o s 3 α}{c o s 4 α c o s 2 α} . 2 \frac{c o s 4 α c o s 2 α}{{c o s}^{2} 3 α {c o s}^{2} α}$ $2 t a n 3 α {s e c}^{2} α$
	Answered by math1967 last updated on 05/Jul/18

	$(\frac{t a n 3 α + t a n α}{1 - \tan 3 α \tan α} + \frac{\tan 3 α - \tan α}{1 + \tan 3 α \tan α}) (1 - {t a n}^{2} 3 α {t a n}^{2} α)$ $\frac{2 t a n 3 α + 2 {t a n}^{2} α t a n 3 α}{(1 - {t a n}^{2} 3 α {t a n}^{2} α)} \times (1 - {t a n}^{2} 3 α {t a n}^{2} α)$ $2 t a n 3 α (1 + {t a n}^{2} α) = 2 t a n 3 α {s e c}^{2} α$
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