Question

use a reference triangle in an appropriate quadrant to find the given angle. cot^(-1)(√3) cot^(-1)(√3)=□ (type an exact answer, using π as needed. use integers or fractions for any number)

Explanation:

Step1: Recall cotangent - inverse relationship

Let $\theta=\cot^{-1}(\sqrt{3})$, then $\cot\theta = \sqrt{3}$. Since $\cot\theta=\frac{\cos\theta}{\sin\theta}$, and we know that in the first - quadrant, we can consider a right - triangle. In a right - triangle, $\cot\theta=\frac{\text{adjacent}}{\text{opposite}}=\sqrt{3}=\frac{\sqrt{3}}{1}$.

Step2: Use special right - triangle values

We know that for a $30 - 60-90$ triangle, if the opposite side to an angle $\theta$ is $1$ and the adjacent side is $\sqrt{3}$, then $\theta$ is an angle such that $\cot\theta=\sqrt{3}$. The angle $\theta$ in the range of the inverse cotangent function $y = \cot^{-1}(x)$ which has a range of $(0,\pi)$ and for $\cot\theta=\sqrt{3}$, $\theta=\frac{\pi}{6}$.

Answer:

$\frac{\pi}{6}$