Let θ be an angle in standard position. Name the quadrant in which the angle θ lies.
-$\cot \theta = - \frac { 3 } { 2 } , \quad \cos \theta

Question

Let θ be an angle in standard position. Name the quadrant in which the angle θ lies.
-$\cot \theta = - \frac { 3 } { 2 } , \quad \cos \theta < 0 \quad$ Find $\csc \theta$
A) $\frac { \sqrt { 13 } } { 2 }$
B) $- \frac { 3 \sqrt { 13 } } { 13 }$
C) $\frac { 3 \sqrt { 13 } } { 13 }$
D) $- \frac { \sqrt { 13 } } { 3 }$

Quizplus · Accepted Answer

The Answer of Let θ be an angle in standard position. Name the quadrant in which the angle θ lies.
-$\cot \theta = - \frac { 3 } { 2 } , \quad \cos \theta < 0 \quad$ Find $\csc \theta$
A) $\frac { \sqrt { 13 } } { 2 }$
B) $- \frac { 3 \sqrt { 13 } } { 13 }$
C) $\frac { 3 \sqrt { 13 } } { 13 }$
D) $- \frac { \sqrt { 13 } } { 3 }$

Let θ Be an Angle in Standard Position cot⁡θ=−32,cos⁡θ<0\cot \theta = - \frac { 3 } { 2 } , \quad \cos \theta < 0 \quadcotθ=−23​,cosθ<0

Let θ Be an Angle in Standard Position $\cot \theta = - \frac { 3 } { 2 } , \quad \cos \theta < 0 \quad$