How do you evaluate \[\sec \left( \dfrac{9\pi }{4} \right)\]?

Hint: From the question, we have been asked to evaluate \[\sec \left( \dfrac{9\pi }{4} \right)\]. We can solve the given question by knowing about trigonometric cycles and basic angle values of trigonometric functions. We know that all trigonometric angles have a periodic cycle of $2\pi $ .

Complete step-by-step solution:
From the question, we have been given that \[\sec \left( \dfrac{9\pi }{4} \right)\]
We already know that all trigonometric ratios have a cycle of \[2\pi \], that is their values repeat after every \[2\pi \].
By using the above written property of trigonometry we can evaluate the given question.
Now, we have to rewrite the given question to make it easier to solve.
\[\sec \left( \dfrac{9\pi }{4} \right)=\sec \left( \dfrac{8\pi }{4}+\dfrac{\pi }{4} \right)\]
By simplifying the above equation furthermore.
\[\sec \left( \dfrac{9\pi }{4} \right)=\sec \left( 2\pi +\dfrac{\pi }{4} \right)\]
As we have been already discussed above,
All trigonometric ratios have a cycle of \[2\pi \], that is their values repeat after every \[2\pi \].
By using the above property, we can rewrite the above equation as,
\[\sec \left( \dfrac{9\pi }{4} \right)=\sec \left( \dfrac{\pi }{4} \right)\]
But, we know the value of \[\sec \left( \dfrac{\pi }{4} \right)=\sqrt{2}\]
Therefore \[\sec \left( \dfrac{9\pi }{4} \right)=\sqrt{2}\]
Hence, the given question is evaluated.
As we have already discussed above, we got the question evaluated by using some basic properties of trigonometric cycles and basic trigonometric angles values.

Note: We should be well aware of the properties of trigonometry. Also, we should know the values of basic angles of trigonometry. Also, we should be very careful while doing the calculation of trigonometric angles and values. Also, we should be well aware of the basic properties of trigonometric cycles. We should know the usage of the properties of the trigonometric cycles. We should be very careful about the usage of the property in the given problem. Similarly the value of $\sin \left( \dfrac{9\pi }{4} \right)$ is given by $\sin \left( 2\pi +\dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ .