[Math] If $\tan t = \frac{1}{4}$ and the terminal point for t is in Quadrant III, find $\sec t + \cot t$.

I have a problem with understanding why my first method of finding trigonometric function is incorrect.

Question:
If $\tan t = \frac{1}{4}$ and the terminal point for $t$ is in Quadrant III, find $\sec t + \cot t$.

Book answer:
According to the book I'm using, the answer is: $\frac{16 – \sqrt{17}}{4}$

Solution 1:
Here's how I tried to solve the problem:

If $\tan t = \frac{\sin t}{\cos t}$ and it's Quadrant III, then $\sin t = -1$ and $\cos t = -4$

If $\tan t = \frac{1}{4}$, then $\cot t$ is $4$

If $\sec t = \frac{1}{\cos t}$, then $\sec t = -\frac{1}{4}$

So, according to my calculations, $\sec t + \cot t = -\frac{1}{4} + 4 = \frac{15}{4}$, but this doesn't seem to be correct.

I also tried to solve the problem using pythagorean identities and I got correct solution:

If, $\tan^2 t + 1 = \sec^2 t$

$\left(\frac{1}{4}\right)^2 + 1 = \sec^2 t\cdot\frac{1}{16} + 1 = \sec^2 t\cdot
\frac{17}{16} = \sec^2 t\cdot \frac{\sqrt{17}}{4} = \sec t$

as it's Quadrant III, $\sec t = – \frac{\sqrt{17}}{4}$

$-\frac{\sqrt{17}}{4} + \frac{16}{4} = \frac{16 – \sqrt{17}}{4}$

I don't understand where things went wrong in Solution 1