The answer is B. Thank me in the future!
always remember ~Positive Vibes~
Answer:
Option B is correct.
Step-by-step explanation:
Given that
[tex]\sec{\theta}=-\frac{37}{12}[/tex]
and the terminal point [tex]\theta[/tex] is in quadrant 2,i.e [tex]\frac{\pi}{2}<\theta<\pi[/tex]
we have to find the value of [tex]\cot \theta[/tex]
As in second quadrant [tex]\cos{\theta}[/tex] is negative.
[tex]\cos{\theta}=\frac{1}{\sec{\theta}}=\frac{1}{\frac{-37}{12}}=-\frac{12}{37}[/tex]
As in second quadrant all trigonometric functions are negative except [tex]\sin{\theta} \thinspace and\thinspace \csc{\theta}[/tex]
[tex]sin{\theta}=\pm\sqrt{1-\cos^2{\theta}}=\pm\sqrt{1-(-\frac{12}{37})^2)}=\pm\sqrt{1-\frac{144}{1369}}=\pm\sqrt{\frac{1225}{1369}}=\frac{35}{37}[/tex]
[tex]\cot{\theta}=\frac{\cos{\theta}}{\sin{\theta}}=\frac{-\frac{12}{37}}{\frac{35}{37}}=-\frac{12}{35}[/tex]
Option B is correct.
