You found the proper value of $\cos \theta$ for the reference angle for $\theta$, not for $\theta$ itself. The reference angle is the angle in the first quadrant that has the same trig values in absolute value. Your final step is to find the proper sign of your trig value, based on the quadrant your $\theta$ is in.
Your question tells you that the tangent is negative, which is true for quadrants II and IV. It also tells you that the sine is positive, which is true for quadrants I and II. The only quadrant that fits both requirements is the second quadrant, so that's where $\theta$ lies. In that quadrant, cosine is negative, so you must change your answer from $\frac{15}{17}$ to $-\frac{15}{17}$.
One way to remember the signs for each quadrant is the mnemonic All Students Take Calculus. This tells you which trig ratios are positive in each quadrant: all, sine, tangent, and cosine for quadrants I, II, III, and IV respectively.
There are other ways, such as trig identities, to solve your problem that give the correct sign automatically, but it is good to have several ways to solve your problem.
Best Answer
Because in math, it is exactly $1$. I have no experience in programming but presumably, when you write a program to approximate $\cos^2\theta+\sin^2\theta$, it will always give something close to the exact value. Here's a short proof. Let $\theta$ be an angle in a right angled triangle, and let the lengths of the adjacent side to that angle, the opposite side, and the hypotenuse be $\mathrm{adj.}$, $\mathrm{opp.}$ and $\mathrm{hyp.}$ respectively.
$$\begin{split}\cos^2\theta+\sin^2\theta & = \left(\frac{\mathrm{adj.}}{\mathrm{hyp}}\right)^2 + \left(\frac{\mathrm{opp.}}{\mathrm{hyp}}\right)^2\\ & = \frac{\mathrm{adj.}^2+\mathrm{opp.}^2}{\mathrm{hyp.}^2}\\ & =^* \frac{\mathrm{hyp.}^2}{\mathrm{hyp.}^2}\\ & = 1. \end{split}$$
Note that the starred equality $=^*$ follows because of the Pythagorean Theorem.