Find the exact value of the indicated trigonometric function of $\theta$
-$\sin \theta = \frac { 1 } { 2 } , \quad \sec \theta

Question

Find the exact value of the indicated trigonometric function of $\theta$
-$\sin \theta = \frac { 1 } { 2 } , \quad \sec \theta < 0$
Find $\cos \theta$ and $\tan \theta$.
A) $\cos \theta = - \sqrt { 3 } , \tan \theta = - \frac { 10 \sqrt { 3 } } { 3 }$
B) $\cos \theta = - \frac { \sqrt { 3 } } { 2 } , \tan \theta = \frac { \sqrt { 3 } } { 3 }$
C) $\cos \theta = - \frac { \sqrt { 3 } } { 2 } , \tan \theta = - \frac { \sqrt { 3 } } { 3 }$
D) $\cos \theta = \sqrt { \frac { 3 } { 2 } } , \tan \theta = \frac { \sqrt { 3 } } { 3 }$

Quizplus · Accepted Answer

Given $\sin \theta = \frac{1}{2}$ and $\sec \theta < 0$, this implies $\theta$ is in the second quadrant where $\sin$ is positive and $\cos$ is negative. The exact value of $\cos \theta$ when $\sin \theta = \frac{1}{2}$ is $-\frac{\sqrt{3}}{2}$, and $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{\sqrt{3}}{3}$.

Find the Exact Value of the Indicated Trigonometric Function Of \theta