Given that $\theta$ is an acute angle of a right triangle and $\sec(\theta) = 2$, we can find $\cot(\theta)$ and $\cos(\theta)$ using trigonometric identities.

Step 1: Find $\cos(\theta)$

We know that the secant function is the reciprocal of the cosine function:

$\sec(\theta) = \frac{1}{\cos(\theta)}$

Given $\sec(\theta) = 2$, we can solve for $\cos(\theta)$:

$\cos(\theta) = \frac{1}{\sec(\theta)} = \frac{1}{2}$

So, $\cos(\theta) = \frac{1}{2}$.

Step 2: Find $\cot(\theta)$

To find $\cot(\theta)$, we need to determine $\sin(\theta)$, and then use the identity $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$.

We can use the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to find $\sin(\theta)$:

$\sin^2(\theta) + \left( \frac{1}{2} \right)^2 = 1$

$\sin^2(\theta) + \frac{1}{4} = 1$

$\sin^2(\theta) = 1 - \frac{1}{4} = \frac{3}{4}$

$\sin(\theta) = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

Now, using $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$:

$\cot(\theta) = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

Final Answer:

• $\cos(\theta) = \frac{1}{2}$
• $\cot(\theta) = \frac{\sqrt{3}}{3}$

Would you like further details or explanations?

Here are 5 related questions for practice:

1. If $\sin(\theta) = \frac{1}{2}$, what is $\sec(\theta)$?
2. Given $\tan(\theta) = \frac{1}{2}$, find $\cot(\theta)$.
3. If $\cos(\theta) = \frac{\sqrt{2}}{2}$, what is $\sin(\theta)$?
4. How do you calculate $\csc(\theta)$ if $\sin(\theta) = \frac{3}{5}$?
5. Given that $\cot(\theta) = 2$, find $\tan(\theta)$.

Tip: Remember, secant and cosine are reciprocal trigonometric functions!