How do I evaluate the trigonometric functions of the quadratic angle? Thanks in advance

1) sec π
2) tan π
3) cos π For this problem, I don't understand why it's -1
4) csc π
5) sec 3π/2

To evaluate the trigonometric functions of a quadratic angle, we need to understand the unit circle and the special angles associated with it.

First, it's important to note that when we mention angles like π (pi), we are referring to radians. In radians, π represents half of the circumference of a circle.

Now, let's evaluate each trigonometric function you mentioned step by step:

1) sec π:
The secant function is the reciprocal of the cosine function. By looking at the unit circle, we can see that at angle π, the x-coordinate of the corresponding point is -1. Therefore, cos π = -1. Now, taking the reciprocal, sec π = 1/(-1) = -1.

2) tan π:
The tangent function is the sine function divided by the cosine function. At angle π, the y-coordinate of the corresponding point is 0, and the x-coordinate is -1. Therefore, sin π = 0 and cos π = -1. So, tan π = sin π / cos π = 0 / (-1) = 0.

3) cos π:
At angle π, the x-coordinate of the corresponding point is -1. Therefore, cos π = -1.

4) csc π:
The cosecant function is the reciprocal of the sine function. At angle π, the y-coordinate of the corresponding point is 0. Therefore, sin π = 0. Now, taking the reciprocal, csc π = 1/0. Since division by zero is undefined, csc π is undefined in this case.

5) sec 3π/2:
Similar to the first question, we need to evaluate the secant function at the angle 3π/2. Looking at the unit circle, we can see that at this angle, the x-coordinate is 0. Therefore, sec 3π/2 = 1/0. Again, division by zero is undefined, so sec 3π/2 is undefined in this case.

Remember, understanding the unit circle and the values of trigonometric functions at special angles is key to evaluating trigonometric functions for any angle.