find the exact value of the trigonometric function at the given real number.\n(a) (sec \frac{11…

Explanation:

Step1: Find the reference angle for (\frac{11\pi}{3})

(\frac{11\pi}{3}- 2\pi\times1=\frac{11\pi - 6\pi}{3}=\frac{5\pi}{3}), (\frac{5\pi}{3}) is in the fourth - quadrant. The reference angle (\theta'=2\pi-\frac{5\pi}{3}=\frac{\pi}{3}) (\sec t=\frac{1}{\cos t}), (\cos\frac{11\pi}{3}=\cos\frac{\pi}{3}=\frac{1}{2}) (\sec\frac{11\pi}{3}=\frac{1}{\cos\frac{11\pi}{3}} = 2)

Step2: Find (\csc\frac{11\pi}{3})

(\csc t=\frac{1}{\sin t}), (\sin\frac{11\pi}{3}=-\sin\frac{\pi}{3}=-\frac{\sqrt{3}}{2}) (\csc\frac{11\pi}{3}=\frac{1}{\sin\frac{11\pi}{3}}=-\frac{2\sqrt{3}}{3})

Step3: Find (\sec(-\frac{\pi}{6}))

Since (\sec(-t)=\sec t) (because (\sec t=\frac{1}{\cos t}) and (\cos(-t)=\cos t)) (\sec(-\frac{\pi}{6})=\sec\frac{\pi}{6}), and (\sec\frac{\pi}{6}=\frac{1}{\cos\frac{\pi}{6}}), (\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}) (\sec(-\frac{\pi}{6})=\frac{2\sqrt{3}}{3})

Answer:

(a) (2) (b) (-\frac{2\sqrt{3}}{3}) (c) (\frac{2\sqrt{3}}{3})