find the exact value of sec(tan^(-1)(-4)).

Explanation:

Step1: Let $\theta=\tan^{-1}(-4)$

This means $\tan\theta = - 4=\frac{-4}{1}$, where $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$.

Step2: Use the identity $1 + \tan^{2}\theta=\sec^{2}\theta$

Substitute $\tan\theta=-4$ into the identity: $1+(-4)^{2}=\sec^{2}\theta$. So, $1 + 16=\sec^{2}\theta$, which gives $\sec^{2}\theta = 17$.

Step3: Determine the sign of $\sec\theta$

Since $\tan\theta=-4<0$ and $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\theta\in(-\frac{\pi}{2},0)$. In this interval, $\sec\theta>0$. So, $\sec\theta=\sqrt{17}$.

Answer:

$\sqrt{17}$