To solve this question, you must know SOHCAHTOA. This acronym can be broken into three parts to solve for the sine, cosine, and tangent.

We can use this information to solve our identity.

Dividing by a fraction is the same as multiplying by its reciprocal. 

The sine divided by cosine is the tangent of the angle.

The cotangent of the angle of a triangle is the adjacent side over the opposite side. The answer is 

Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.")  

Now, it is easiest to think of this like you are drawing a little triangle in the third quadrant of the Cartesian plane. It would look like: 

So, the tangent of an angle is:

  or, for your data, . 

This is . Rounding, this is .  Since  is in the third quadrant, your value must be positive, as the tangent function is positive in that quadrant.

Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.")  

Now, it is easiest to think of this like you are drawing a little triangle in the third quadrant of the Cartesian plane. It would look like: 

So, the tangent of an angle is:

  or, for your data, , or . Since  is in the third quadrant, your value must be positive, as the tangent function is positive in this quadrant.

Based on our information, we can draw this little triangle:

Since we know that the tangent of an angle is , we can say:

This can be solved using your calculator:

 or 

Now, to solve for , use the Pythagorean Theorem, , where  and  are the legs of a triangle and  is the triangle's hypotenuse. Here, , so we can substitute that in and rearrange the equation to solve for :

Substituting in the known values:

, or approximately . Rounding, this is .

You can begin by imagining a little triangle in the second quadrant to find your reference angle. It would look like this:

The tangent of an angle is:

For our data, this is:

Now, since this is in the second quadrant, the value is negative, given the periodic nature of the tangent function.

Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.")  

Now, it is easiest to think of this like you are drawing a little triangle in the second quadrant of the Cartesian plane. It would look like: 

So, the tangent of an angle is:

  or, for your data, .

This is . Rounding, this is . However, since  is in the second quadrant, your value must be negative. (The tangent function is negative in that quadrant.) Therefore, the answer is .

Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the -axis as your reference point for your angle. (Hence, it is called the "reference angle.")  

Now, it is easiest to think of this like you are drawing a little triangle in the fourth quadrant of the Cartesian plane. It would look like:

So, the tangent of an angle is:

  or, for your data,  or . However, since  is in the fourth quadrant, your value must be negative. (The tangent function is negative in that quadrant.) This makes the correct answer .

The standard period of a tangent function is  radians. In other words, it completes its entire cycle of values in that many radians. To alter the period of the function, you need to alter the value of the parameter of the trigonometric function. You multiply the parameter by the number of periods that would complete in  radians. With a period of , you are quadrupling your method. Therefore, you will have a function of the form:

Since  and  do not alter the period, these can be anything.

Therefore, among your options,  is correct.

The standard period of a tangent function is  radians. In other words, it completes its entire cycle of values in that many radians. To alter the period of the function, you need to alter the value of the parameter of the trigonometric function. You multiply the parameter by the number of periods that would complete in  radians. With a period of , you are multiplying your parameter by . Now, half of this would be a period of . Thus, you will have a function of the form:

Since  and  do not alter the period, these can be anything.

Therefore, among your options,  is correct.