We are given the hypotenuse and the side opposite of the angle in question. The trig function that relates these two sides is SIN. Therefore, we can write:

$\displaystyle sin(A) = \frac{12}{22}$

In order to solve for A, we need to take the inverse sin of both sides:

$\displaystyle sin^{-1}(sin(A)) = sin^{-1}(\frac{12}{22})$

which becomes

$\displaystyle A = sin^{-1}\bigg(\frac{12}{22}\bigg) = 33^{\circ}$

Solve for theta by taking the inverse sine of both sides.

$\displaystyle \theta=sin^-^1\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$

Since this angle is not valid for the given interval of theta, add $\displaystyle 2\pi$ radians to this angle to get a valid answer in the interval.

$\displaystyle -\frac{\pi}{6}+2\pi = -\frac{\pi}{6}+\frac{12\pi}{6}= \frac{11\pi}{6}$

First evaluate $\displaystyle cos^-^1\left(-\frac{\sqrt3}{2}\right)$.

To evaluate inverse cosine, it is necessary to know the domain and range of inverse cosine. 

For: $\displaystyle \theta=cos^-^1(x)$

The domain $\displaystyle x$ is only valid from $\displaystyle [-1,1]$.

$\displaystyle \theta$ is only valid from $\displaystyle [0,\pi]$.

The part is asking for the angle where the x-value of the coordinate is $\displaystyle -\frac{\sqrt3}{2}$.  The only possibility on the unit circle is the second quadrant.  

$\displaystyle cos^-^1\left(-\frac{\sqrt3}{2}\right)= \frac{5}{6}\pi$

Next, evaluate $\displaystyle cos^-^1\left(\frac{\sqrt3}{2}\right)$.

Using the same domain and range restrictions, the only valid angle for the given x-value is in the first quadrant on the unit circle.  

$\displaystyle cos^-^1\left(\frac{\sqrt3}{2}\right)= \frac{\pi}{6}$

Therefore:

$\displaystyle cos^-^1\left(-\frac{\sqrt3}{2}\right)+cos^-^1\left(\frac{\sqrt3}{2}\right) = \frac{5}{6}\pi + \frac{\pi}{6}=\frac{6\pi}{6}=\pi$

To find the correct value of $\displaystyle cos^-^1\left(\frac{1}{2}\right)$, it is necessary to know the domain and range of inverse cosine.

Domain:  $\displaystyle [-1,1]$

Range:  $\displaystyle [0,\pi]$

The question is asking for the specific angle when the x-coordinate is half.  

The only possibility is located in the first quadrant, and the point of the special angle is $\displaystyle \left(\frac{1}{2},\frac{\sqrt3}{2}\right)$

The special angle for this coordinate is $\displaystyle \frac{\pi}{3}$.

In order to determine the value or values of $\displaystyle cos^-^1(\frac{1}{2})$, it is necessary to know the domain and range of the inverse sine function.

Domain:  $\displaystyle [-1,1]$

Range:  $\displaystyle [0,\pi]$

The question is asking for the angle value of theta where the x-value is $\displaystyle \frac{1}{2}$ under the range restriction.  Since $\displaystyle x=\frac{1}{2}$ is located in the first and fourth quadrants, the range restriction makes theta only allowable from $\displaystyle [0,\pi]$.  Therefore, the theta value must only be in the first quadrant.

The value of the angle when the x-value is $\displaystyle \frac{1}{2}$ is $\displaystyle 60$ degrees.

The easiest way to solve this problem is to simplify the original expression. 

$\displaystyle y=csc(x)*tan(x)=\frac{1}{sin(x)}*\frac{sin(x)}{cos(x)}= \frac{1}{cos(x)}$

$\displaystyle y=\frac{1}{cos(x)}$

To find its inverse, let's exchange $\displaystyle x$ and $\displaystyle y$, 

$\displaystyle x=\frac{1}{cos(y)}$

Solving for $\displaystyle y$

$\displaystyle \frac{1}{x}=cos(y)$

$\displaystyle arccos(\frac{1}{x})=y$