First, we find an angle that's coterminal with $\dfrac{9\pi}{2}$ that lies in the range $[0,2\pi).$

\begin{align} \dfrac{9\pi}{2} - 2\pi &= \dfrac{5\pi}{2}\\[5pt] \dfrac{5\pi}{2} - 2\pi &= \dfrac{\pi}{2}\quad{\color{green}{\checkmark}} \end{align}

Therefore, $\cos \left(\dfrac{9\pi}2\right) = \cos \left(\dfrac{\pi}2\right).$

The angle $\dfrac{\pi}{2}$ is a quadrantal angle. Let's draw out the quadrantal angles on the unit circle.

From the unit circle, we see that the angle $\dfrac{\pi}{2}$ corresponds to the point $(0,1).$

Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos{\left(\dfrac{\pi}{2}\right)} = 0.$

Therefore, $\cos \left(\dfrac{9\pi}2\right) = 0.$

First, we find an angle that's coterminal with $2\pi$ that lies in the range $[0,2\pi).$

\[ 2\pi - 2\pi = 0\quad{\color{green}{\checkmark}} \]

Therefore, $\cos 0 = \cos 2\pi.$

The angle $0$ is a quadrantal angle. Let's draw out the quadrantal angles on the unit circle.

From the unit circle, we see that the angle $0$ corresponds to the point $(1,0).$

Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos 0 = 1.$

Therefore, using the fact that $\sec{\theta} = \dfrac{1}{\cos\theta},$ we have

\[ \sec{0} = \dfrac{1}{\cos{0} }= \dfrac{1}{1}. \]

Therefore, $\sec 2\pi = 1.$

First, we find an angle that's coterminal with $-270^\circ$ that lies in the range $[0^\circ, 360^\circ).$

\[ -270^\circ + 360^\circ = 90^\circ\quad{\color{green}{\checkmark}} \]

Therefore, $\sec\left(-270^\circ\right) = \sec{90^\circ}.$

The angle $90^\circ$ is a quadrantal angle. Let's draw out the quadrantal angles on the unit circle.

From the unit circle, we see that the angle $90^\circ$ corresponds to the point $(0,1).$

Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos{90^\circ} = 0.$

Therefore, using the fact that $\sec{\theta} = \dfrac{1}{\cos\theta},$ we have

\[ \sec{(90^\circ)} = \dfrac{1}{\cos{(90^\circ)}} = \dfrac{1}{0}. \]

However, division by zero is undefined. Therefore, since $\sec(-270^\circ)$ is undefined, $\sec(90^\circ)$ is also undefined.

First, we find an angle that's coterminal with $-90^\circ$ that lies in the range $[0,360^\circ).$

\[ -90^\circ + 360^\circ = 270^\circ\quad{\color{green}{\checkmark}} \]

Therefore, $\csc \left(-90^\circ \right) = \csc \left(270^\circ\right).$

The angle $270^\circ$ is a quadrantal angle. Let's draw out the quadrantal angles on the unit circle.

From the unit circle, we see that the angle $270^\circ$ corresponds to the point $(0,-1).$

Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin\left(270^\circ\right) = -1.$

Therefore, using the fact that $\csc{\theta} = \dfrac{1}{\sin\theta},$ we have

\[ \csc \left(270^\circ \right) = \dfrac{1}{\sin{\left(270^\circ \right) }} = \dfrac{1}{-1} = -1. \]

Therefore, $\csc \left(-90^\circ \right) = -1.$