tan xtan z=3 tan ytan z=6 x+y+z = π Solve for x, y, and z.


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Question Number 20721 by ajfour last updated on 01/Sep/17

$\tan x \tan z = 3$ $\tan y \tan z = 6$ $x + y + z = π$ $S o l v e f o r x, y, a n d z .$
	Answered by sma3l2996 last updated on 02/Sep/17

	$t a n (x + y + z) = \frac{t a n (x + y) + t a n (z)}{1 - t a n (x + y) t a n (z)} = 0$ $= \frac{\frac{t a n x + t a n y}{1 - t a n x t a n y} + t a n z}{1 - \frac{t a n x + t a n y}{1 - t a n x t a n y} t a n z} = \frac{t a n x + t a n y + t a n z - t a n x t a n y t a n z}{1 - (t a n x t a n y + t a n x t a n z + t a n y t a n z)} = 0$ $t a n x + t a n y + t a n z - t a n x t a n z t a n y t a n z \times \frac{1}{t a n z} = 0$ $\frac{3}{t a n z} + \frac{6}{t a n z} + t a n z - \frac{18}{t a n z} = 0$ $\frac{1}{t a n z} ({t a n}^{2} z - 9) = 0 \Leftrightarrow t a n z = 3 o r t a n z = - 3$ $z = {t a n}^{- 1} (3) + \frac{k π}{2}$ $s o t a n x = \frac{3}{t a n z} = \frac{3}{3} = 1 o r t a n x = - 1$ $x = \frac{π}{4} + \frac{k π}{2} a n d y = {t a n}^{- 1} (2) + \frac{k π}{2}$
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