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 \(\displaystyle y\)

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

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