tan-xtan-z-3-tan-ytan-z-6-x-y-z-pi-Solve-for-x-y-and-z-

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}