Answer:
[tex] \tan(x) = \frac{ -\sqrt{2} }{4} [/tex]
Step-by-step explanation:
So an angle has two parts. Initial side and terminal side.
Inital side like on x axis. and terminal side shows how much it open up. Here the terminal angle terminates in second quadrant so we have the following
Now, using Pythagoras identity let solve for cos theta.
Here you on the right track but remeber that son theta=1/3 so sin theta squared would be 1/3 squared so we have
[tex] (\frac{1}{3} ) {}^{2} + \cos {}^{2} (x) = 1[/tex]
[tex] \frac{1}{9} + \cos {}^{2} (x) = 1[/tex]
[tex] \cos {}^{2} (x) = \frac{8}{9} [/tex]
[tex] \cos(x) = \frac{2 \sqrt{2} }{3} [/tex]
Note since cosine is negative in second quadrant, cos theta is
[tex] - \frac{2 \sqrt{2} }{3} [/tex]
To find tan theta we do the following
[tex] \tan(x) = \frac{ \sin(x) }{ \cos(x) } [/tex]
[tex] \tan(x) = \frac{ \frac{1}{3} }{ \frac{2 \sqrt{2} }{3} } [/tex]
[tex] \tan(x) = \frac{1}{2 \sqrt{2} } [/tex]
[tex] \tan(x) = \frac{2 \sqrt{2} }{8} [/tex]
[tex] \frac{ \sqrt{2} }{4} [/tex]
So
[tex] \tan(x) = \frac{ -\sqrt{2} }{4} [/tex]
Tan is negative in second quadrant