Question

if $sec(\theta)=\frac{13}{5}$ and $0^{circ}<\theta<90^{circ}$, what is $cot(\theta)$? write your answer in simplified, rationalized form. $cot(\theta)=$

Explanation:

Step1: Recall the definition of secant

Since $\sec(\theta)=\frac{13}{5}$, and $\sec(\theta)=\frac{1}{\cos(\theta)}$, then $\cos(\theta)=\frac{5}{13}$.

Step2: Use the Pythagorean identity

We know that $\sin^{2}\theta+\cos^{2}\theta = 1$. Substituting $\cos(\theta)=\frac{5}{13}$, we get $\sin^{2}\theta=1 - (\frac{5}{13})^{2}=1-\frac{25}{169}=\frac{144}{169}$. Since $0^{\circ}<\theta<90^{\circ}$, $\sin(\theta)=\frac{12}{13}$.

Step3: Recall the definition of cotangent

$\cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}$. Substituting $\cos(\theta)=\frac{5}{13}$ and $\sin(\theta)=\frac{12}{13}$, we have $\cot(\theta)=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}$.

Answer:

$\frac{5}{12}$