Let θ be an angle in standard position. Name the quadrant in which the angle θ lies.
-$\sin \theta = - \frac { 4 } { 9 } , \quad \tan \theta 0 \quad$ Find $\sec \theta$
A) $- \frac { 9 \sqrt { 65 } } { 65 }$
B) $- \frac { \sqrt { 65 } } { 9 }$
C) $\frac { \sqrt { 9 } } { 4 }$
D) $- \frac { 4 \sqrt { 65 } } { 65 }$

Question

Let θ be an angle in standard position. Name the quadrant in which the angle θ lies.
-$\sin \theta = - \frac { 4 } { 9 } , \quad \tan \theta > 0 \quad$ Find $\sec \theta$
A) $- \frac { 9 \sqrt { 65 } } { 65 }$
B) $- \frac { \sqrt { 65 } } { 9 }$
C) $\frac { \sqrt { 9 } } { 4 }$
D) $- \frac { 4 \sqrt { 65 } } { 65 }$

Quizplus · Accepted Answer

The Answer of Let θ be an angle in standard position. Name the quadrant in which the angle θ lies.
-$\sin \theta = - \frac { 4 } { 9 } , \quad \tan \theta > 0 \quad$ Find $\sec \theta$
A) $- \frac { 9 \sqrt { 65 } } { 65 }$
B) $- \frac { \sqrt { 65 } } { 9 }$
C) $\frac { \sqrt { 9 } } { 4 }$
D) $- \frac { 4 \sqrt { 65 } } { 65 }$

Let θ Be an Angle in Standard Position sin⁡θ=−49,tan⁡θ>0\sin \theta = - \frac { 4 } { 9 } , \quad \tan \theta > 0 \quadsinθ=−94​,tanθ>0

Let θ Be an Angle in Standard Position $\sin \theta = - \frac { 4 } { 9 } , \quad \tan \theta > 0 \quad$