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 sine of the angle corresponds to the $y$-coordinate, we have $\sin({90^\circ}) =\bbox[4pt,border: 1px solid lightgray]{ 1}.$

The angles $0^\circ$ and $270^\circ$ are quadrantal angles. Let's draw out the quadrantal angles on the unit circle.

• Let's first consider the angle $0^\circ.$ From the unit circle, we see that the angle $0^\circ$ corresponds to the point $(1,0).$ Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos({0^\circ}) = 1.$

• Now, let's consider the angle $270^\circ.$ 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 $\sin({270^\circ}) = -1.$

Finally, we get \[ \cos({0^\circ}) + \sin({270^\circ}) = 1 + (-1) = 0. \]

The angle $180^\circ$ is a quadrantal angle. Let's draw out the quadrantal angles on the unit circle.

From the unit circle, we see that the angle $180^\circ$ corresponds to the point $(-1,0).$

• Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos({180^\circ}) = -1.$

• Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin({180^\circ}) = 0.$

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

\[ \tan{180^\circ} = \dfrac{\sin({180^\circ})}{\cos({180^\circ})} = \dfrac{0}{-1}. \]

Therefore, $\tan(180^\circ)=\bbox[4pt,border: 1px solid lightgray]{0}.$

The angles $180^\circ$ and $90^\circ$ are quadrantal angles. Let's draw out the quadrantal angles on the unit circle.

Let's first consider the angle $180^\circ.$ From the unit circle, we see that the angle $180^\circ$ corresponds to the point $(-1,0).$

• Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos({180^\circ}) = -1.$

• Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin({180^\circ}) = 0.$

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

\[ \tan({180^\circ}) = \dfrac{\sin({180^\circ})}{\cos({180^\circ})} = \dfrac{0}{-1}=0. \]

Now, let's consider the angle $90^\circ.$ 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 $\cos({90^\circ}) = 0.$

• Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin{({90^\circ})} = 1.$

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

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

Finally, we get \[ \tan (180^\circ) + \cot({90^\circ}) = 0 + 0 =\bbox[4pt,border: 1px solid lightgray]{ 0}. \]

The angle $\dfrac{3\pi}{2}$ is a quadrantal angle. Let's draw out the quadrantal angles on the unit circle.

From the unit circle, we see that the angle $\dfrac{3\pi}{2}$ corresponds to the point $(0,-1).$

• Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos\left( \dfrac{3\pi}{2}\right) = 0.$

• Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin\left( \dfrac{3\pi}{2}\right) = -1.$

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

\[ \tan{\left( \dfrac{3\pi}{2}\right)} = \dfrac{\sin{\left( \dfrac{3\pi}{2}\right) }}{\cos{\left( \dfrac{3\pi}{2}\right) }} = \dfrac{-1}{0}. \]

However, division by zero is undefined. Therefore, $\tan\left( \dfrac{3\pi}{2}\right)$ is undefined.

The angle $\dfrac{3\pi}{2}$ is a quadrantal angle. Let's draw out the quadrantal angles on the unit circle.

From the unit circle, we see that the angle $\dfrac{3\pi}{2}$ corresponds to the point $(0,-1).$

Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin\left(\dfrac{3\pi}{2}\right) = -1.$

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

\[ \csc\left(\dfrac{3\pi}{2}\right) = \dfrac{1}{\sin\left(\dfrac{3\pi}{2}\right)} = \dfrac{1}{-1} = -1. \]

Therefore, $\csc \left(\dfrac{3\pi}{2}\right)=-1.$

The angles $0^\circ$ and $90^\circ$ are quadrantal angles. Let's draw out the quadrantal angles on the unit circle.

• Let's first consider the angle $0^\circ.$ From the unit circle, we see that the angle $0^\circ$ corresponds to the point $(1,0).$ Since the cosine of the angle corresponds to the $x$-coordinate, we have $\cos(0^\circ) = 1.$ Therefore, using the fact that $\sec{\theta} = \dfrac{1}{\cos\theta},$ we have \[ \sec(0^\circ) = \dfrac{1}{\cos(0^\circ)} = \dfrac{1}{1}=1. \]

• Now, let's consider the angle $90^\circ.$ From the unit circle, we see that the angle $90^\circ$ corresponds to the point $(0,1).$ Since the sine of the angle corresponds to the $y$-coordinate, we have $\sin(90^\circ) = 1.$ Therefore, using the fact that $\csc{\theta} = \dfrac{1}{\sin\theta},$ we have \[ \csc(90^\circ) = \dfrac{1}{\sin(90^\circ)} = \dfrac{1}{1}=1. \]

Finally, we get \[ \sec(0^\circ) + \csc(90^\circ) = 1 + 1 = \bbox[4pt,border: 1px solid lightgray]{2}. \]