How do you evaluate $\sec \left( \dfrac{11\pi }{3} \right)$ ?
(a) Using linear formulas
(b) Using trigonometric identities
(c) Using algebraic properties
(d) None of these

Hint: In the given problem, we are trying to find the value of $\sec \left( \dfrac{11\pi }{3} \right)$. We start with writing $\dfrac{11\pi }{3}$as $\dfrac{7\pi }{2}+\dfrac{\pi }{6}$. Hence, we check on which quadrant the angle is and then using the trigonometric table we get our needed solution.

Complete step by step solution:
According to the question, we are trying to evaluate the value of $\sec \left( \dfrac{11\pi }{3} \right)$.
We are starting with the given value, $\dfrac{11\pi }{3}$
Now, $\dfrac{11\pi }{3}$can be written as, $\dfrac{7\pi }{2}+\dfrac{\pi }{6}$ .
Again, as we now, $\pi $ radian can be written as $180{}^\circ $.
So, $\dfrac{7\pi }{2}$can be written as, $\left( \dfrac{7}{2}\times 180 \right){}^\circ =630{}^\circ $
But, also, $630{}^\circ $can be written as a sum of $540{}^\circ $and $90{}^\circ $.
Using the first $540{}^\circ $we are back in the second quadrant. And the next $90{}^\circ $gives us the value in the third quadrant.
As it is also going forward from the third quadrant, we get our angle in our fourth quadrant.
Again, from the all-sin-tan-cos rule, we can easily conclude that the value of the angle in that given quadrant will be positive.
So, from, $\sec \left( \dfrac{11\pi }{3} \right)$ we are getting, $\sec \left( \dfrac{7\pi }{2}+\dfrac{\pi }{6} \right)$ .
From this we have, $\sec \left( \dfrac{7\pi }{2}+\dfrac{\pi }{6} \right)=\sec \left( \dfrac{\pi }{6} \right)$.
And again, $\pi $ radian can be written as $180{}^\circ $.
So, from this, $\dfrac{\pi }{6}$ can be written as, $\dfrac{\pi }{6}=\left( \dfrac{1}{6}\times 180 \right){}^\circ =30{}^\circ $
Thus, we get, $\sec \left( \dfrac{\pi }{6} \right)=\sec 30{}^\circ $
From the trigonometric table identities, we get now, $\sec \left( \dfrac{\pi }{6} \right)=\sec 30{}^\circ =\dfrac{1}{\cos 30{}^\circ }=\dfrac{1}{\dfrac{\sqrt{3}}{2}}=\dfrac{2}{\sqrt{3}}$
Hence, we are having the value of $\sec \left( \dfrac{11\pi }{3} \right)$as, $\dfrac{2}{\sqrt{3}}$ .
So, the solution is, (b) Using trigonometric identities.

Note: ASTC rule is nothing but "all sin tan cos" rule in trigonometry. The angles which lie between 0° and 90° are said to lie in the first quadrant. The angles between 90° and 180° are in the second quadrant, angles between 180° and 270° are in the third quadrant and angles between 270° and 360° are in the fourth quadrant. In the first quadrant, the values for sin, cos and tan are positive. In the second quadrant, the values for sin are positive only. In the third quadrant, the values for tan are positive only. In the fourth quadrant, the values for cos are positive only.