My way of doing this kind of problem usually differs from the textbook methods. I treat the numerical component and the +/- component separately. For the numerical component, I ignore any negative signs for the moment and just place the parts of the fraction where they would go on a right triangle. I deal with the +/- part later.
So if cot θ = -12/5, then (ignoring the negative) that gives us a right triangle with opposite side 5 and adjacent side 12. Both some and secant require a hypotenuse measure, so I can get that by Pythagorean theorem. But you might also recognize these numbers as being part of a very common Pythagorean triple, namely the 5-12-13 triangle. In any case, the hypotenuse will be 13.
Now we have what we need for the numbers in the fractions. So, on this right triangle, sin θ = 5/13, and sec θ = 13/12.
But, of course, this problem didn't start out on some right triangle floating out there in some Euclidean plane. It's on a coordinate plane, with positive and negative numbers. It's in quadrant IV, where x values are positive and y values negative. If you remember the unit circle, cosines are the x values, and sines are the y values.
So, in this case, since sines are negative (since sines are y values), then our sine value will be sin θ = -5/13. However, cosines are positive, and therefore, secants (which are reciprocals of cosines) are also positive. So this will stay sec θ = 13/12.
Heer Z.
Actually the answer was sin 0 = -5/13 and sec 0 = 13/1211/17/19