Precalculus, Chapter 5, 5.4, Section 5.4, Problem 45

There are a lot of ways to solve this problem. For instance, one can use the secant sum/difference identity. This identity, however, has a lot of terms. A more elegant solution would be to use the cosine sum/difference identity, and relating cosine to secant.
Note that sec(x) = 1/(cos(x)) .Such that we only need to get the reciprocal of the value of cosine at our given angle to get the corresponding secant. Both u and v are in the second quadrant implying that the x-components are both negative and the y-components positive.
The cosine sum/difference identity is: cos(v-u) = cos(v)cos(u) + sin(v)sin(u) .
We already know some of the terms here, namely: sin(u) = 5/13 and cos(v) = -3/5 .To get the remaining terms, we simply use the Pythegorean identity: sin^2(x) + cos^2(x) = 1 .
Hence:
cos^2(u) = 1 - 25/169 = 144/169 , and cos(u) = -12/13 . Note that we could have chosen the positive root, but since u is in the second quadrant, we get the negative one.
Similarly:
sin^2(v) = 1 - 9/25 = 16/25 , and sin(v) = 4/5 . We are getting the positive root since v is in the second quadrant.
Now, we can solve for the cosine:
cos(v-u) = (-3/5)(-12/13) + (4/5)(5/13) = 36/65 + 20/65 = 56/65.
Which means:
sec(v-u) = 65/56.