Honors pre-calc anybody? (or trig).
also, do a great job explaining how you got the answer and why.

(i'll give brainliest to anybody who can explain how to do any other problems like this in general).

  sec(θ) = -5/2

Step-by-step explanation:

It can be helpful to draw a triangle with sides that produce the given ratio. See the attachment. Then the third side of the triangle can be found using the Pythagorean theorem, if necessary. Finally, the ratio for the desired trig function can be determined.

  sec = hypotenuse/adjacent

  sec(θ) = 5/-2 = -5/2

__

If you don't want to do that, or if you have various trig identities memorized, you can use the appropriate trig identity.

  sin² + cos² = 1

so ...

  sec² = 1/cos² = 1/(1 -sin²)

  sec = 1/√(1 -sin²)

Since the tangent is positive, this is a third-quadrant angle. The secant will be negative.

  sec(θ) = -1/√(1 -(21/25)) = -√(25/4) . . . . . filling in the given value for sin(θ)

  sec(θ) = -5/2

idk

step-by-step explanation: i just don't