Answer
$1$
Work Step by Step
The reference angle of an angle $0 \leq \theta \lt 2\pi $ based on its position can be computed by using the following steps:
a) Quadrant- I: $\theta $
b) Quadrant -II: $(\pi-\theta)$
c) Quadrant- III: $(\theta - \pi)$
d) Quadrant - IV: $(2\pi - \theta)$
First, we reduce the angle by $2\pi$:
$\dfrac{13 \pi}{4}- \dfrac{8 \pi}{4}=\dfrac{5\pi}{4}$
Since we are in quadrant 3, we subtract $\pi$:
$\dfrac{5\pi}{4}-\pi=\dfrac{5\pi}{4}-\dfrac{4\pi}{4}=\dfrac{\pi}{4}$
Thus, we have $\tan \dfrac{\pi}{4}=1$
The answer is positive because tangent is positive in quadrant 3.
