Answer:
[tex]\dfrac{5}{3}[/tex]
Step-by-step explanation:
We know that [tex]\sec{\theta}=\dfrac{1}{\cos{\theta}}[/tex]. On the other hand,
we can use the equality
[tex]\sin^2 \theta + \cos^2 \theta=1[/tex]
to find the absolute value of [tex]\cos \theta[/tex]. It holds that
[tex]\left[\dfrac{4}{5}\right]^2+\cos^2 \theta = 1 \Rightarrow \\\\\\\quad \cos^2 \theta = 1-\dfrac{16}{25}=\dfrac{9}{25} \Rightarrow\\ \\\cos \theta = \pm \dfrac{3}{5}[/tex]
Now, in the quadrant 4 the values of cosine is positive for all theta, hence
[tex]\cos{\theta}=\dfrac{3}{5} \quad \Rightarrow \quad \sec \theta=\dfrac{1}{\cos\theta}=\dfrac{5}{3}[/tex]
