Answer
$\dfrac{16-\sqrt{17}}{4}$
Work Step by Step
Given: $\tan t=\dfrac{1}{4}$
This implies that
$\cot t=\dfrac{1}{\tan t}=4$ and
$\sec^2 t=1+\tan^2 t \implies \sec^2 t=1+(\dfrac{1}{4})^2$
or, $\sec^2 t=\dfrac{17}{16}$
As $t$ lies in the third quadrant, thus the value of $\sec$ will be negative.
That is, $\sec t=-\sqrt {\dfrac{17}{16}}=-\dfrac{\sqrt {17}}{4}$
Now,
$\sec t+\cot t=-\dfrac{\sqrt {17}}{4}+4=\dfrac{16-\sqrt{17}}{4}$
