The Solution:
Given:
[tex]\begin{gathered} \sin\theta=-\frac{1}{3} \\ \\ \tan\theta>0 \end{gathered}[/tex]We are asked to find the value of:
[tex]\cos\theta[/tex]So,
[tex]\begin{gathered} \sin\theta=-\frac{1}{3} \\ \\ Taking\text{ the }\sin^{-1}\text{ of both sides, we get:} \end{gathered}[/tex][tex]\theta=\sin^{-1}(-\frac{1}{3})=-19.47^o[/tex]Note:
Sins of angles is only negative in the 3rd and 4th quadrants.
Since, it is given that:
[tex]\tan\theta>0\text{ \lparen positive\rparen}[/tex]Thus, it follows that the angle of concern falls into the 3rd quadrant.
