Looking at our graph, we can tell that the period is .  Using the formula 

 where  is the coefficient of  and  is the period, we can calculate 

This eliminates one answer choice.  We then retrun to our graph and see that the amplitude is 3.  Remembering that the amplitude is the number in front of the function, we can eliminate two more choices.

We then examine our graph and realize it contains the point .  Plugging 0 into our two remaining choices, we can determine which one gives us 4 for a result.

 is shown in red, and  is shown in blue.

 In order to graph , recall that .  First consider the graph .

Now anywhere this graph crosses the x-axis a vertical asymptote will form for the  graph because the denominator of  will be equal to zero and the function will be undefined.  At each maximum and minimum of , the graph of  will invert at that point.

And then we are left with the graph of  .

Since cosecant is a reciprocal of sine, it uses the same general formula of the sine function with the letters corresponding to the same transformations.  Note that while A does correspond to amplitude, the cosecant function extends infinitely upwards and downwards so there is no amplitude for the graphs.

Knowing that the graph of  is

we can use the general form of the cosecant transformation equation, , and apply these transformations.  We can ignore  because in this case .  In this case  and so our period is: 

Period = 

Period = 

Period = 

 is the normal period for cosecant graphs and so we do not have to worry about lengthening or shortening the period.   and so we need to apply a phase shift of .  This will cause our graph to shift left a total of  units.

Lastly, we must apply the transformation for , so we will have an upward vertical shift of 1 unit.

The application of these transformations leaves us with our graph of  . 

 In order to understand the graph of secant, recall that .  First consider the graph of .

Anywhere this graph crosses the x-axis a vertical asymptote will form for the  graph because the denominator of  will be equal to zero and the function will be undefined.  At each maximum and minimum of , the graph of  will invert at that point.

And then we are left with the graph . 

When looking at the graph of , it extends infinitely upwards and downwards from each local maximum and minimum.  This will be true for all transformed secant graphs as well.  Due to this, there is no amplitude for secant graphs.  However, secant is the reciprocal of cosine graphs which do rely on amplitude for transformations.  For this reason amplitude must be considered as a vertical shift.

Knowing that the general form of the graph  is:

We can use the general form of the cosecant transformation equation, , and apply these transformations.   because secant graphs extend infinitely upwards and downwards and does not have an amplitude, we must think of the secant graph being a reciprocal of the cosine graph.  So we will consider  for cosine.

We will shift our secant graph to invert at the maximums and minimums of the cosine graph.

Next, we will factor  in order to get our equation into the form .

And so .  We can now solve for our period,

Period = 

Period = 

Period = 

This shortens our original period of  to .

Now we must consider .  This will give us a phase shift of  units to the left.  Since our period has also been shortened this does not change the graph visually. in this case so we do not need to consider a vertical shift.

And we are left with the graph of .

In trigonometry,

.

It may also be thought of as .

This is because

and , so

. 

This means that whenever cosine is 0, tangent is undefined, because it would be evaluated by dividing by 0.

We can factor the original expression as follows:

So from this equation we conclude either that:

or

So any number that is not some integer multiple of  away from these two solutions is not a solution to the original equation.

The only such choice is  , which is  ; n is not an integer, therefore it is not a solution.