which of the following angle(s) in standard position (in radians), if any, is $sec(\\theta)=-\\frac{\\sqrt{2}…

which of the following angle(s) in standard position (in radians), if any, is $sec(\\theta)=-\\frac{\\sqrt{2}}{2}$?\n$\\frac{3\\pi}{4}\\&\\frac{5\\pi}{4}$\nno such angle(s) in standard position exist.\n$\\frac{\\pi}{4}\\&\\frac{7\\pi}{4}$\n$\\frac{\\pi}{4}\\&\\frac{3\\pi}{4}$
Answer
Explanation:
Step1: Recall the secant - cosine relationship
We know that (\sec\theta=\frac{1}{\cos\theta}). Given (\sec\theta =-\frac{\sqrt{2}}{2}), then (\cos\theta=\frac{1}{\sec\theta}=-\sqrt{2}).
Step2: Analyze the range of cosine function
The range of the cosine function is ([- 1,1]). Since (-\sqrt{2}\approx - 1.414<-1), there is no real - valued angle (\theta) (in standard position) for which (\cos\theta =-\sqrt{2}).
Answer:
No such angle(s) in standard position exist.