Let me analyze the image to solve your question. It looks like you’re asking about rationalizing $\frac{1}{\sqrt{15}}$. I’ll confirm by processing the problem from the image and providing clarification. One moment!From the provided image, here's the breakdown of the problem:

Given Information:

The triangle is described with the cosine ratio $\cos B = \frac{1}{4}$. This implies the adjacent side to angle $B$ is 1, and the hypotenuse is 4 (as cosine is defined as $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$).

We are tasked with finding the remaining trigonometric ratios.

Step 1: Solve for the missing side

Using the Pythagorean theorem:

$\text{Hypotenuse}^2 = \text{Adjacent}^2 + \text{Opposite}^2$ $4^2 = 1^2 + \text{Opposite}^2$ $16 = 1 + \text{Opposite}^2$ $\text{Opposite}^2 = 15$ $\text{Opposite} = \sqrt{15}$

Step 2: Calculate the trigonometric ratios

1. Sine: $\sin B = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\sqrt{15}}{4}$
2. Tangent: $\tan B = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sqrt{15}}{1} = \sqrt{15}$
3. Cosecant: $\csc B = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{4}{\sqrt{15}}$. Rationalizing: $\csc B = \frac{4}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}} = \frac{4\sqrt{15}}{15}$
4. Secant: $\sec B = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{4}{1} = 4$
5. Cotangent: $\cot B = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{1}{\sqrt{15}}$. Rationalizing: $\cot B = \frac{1}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}} = \frac{\sqrt{15}}{15}$

Answer to Your Question

Yes, you are correct: for cotangent, the value starts as $\frac{1}{\sqrt{15}}$. When rationalized, it becomes $\frac{\sqrt{15}}{15}$.

Would you like further clarification or detailed steps on any of these calculations?

Related Questions

1. What is the geometric interpretation of rationalizing the denominator?
2. Why is rationalizing necessary in trigonometric problems?
3. How do the trigonometric ratios relate to real-world measurements?
4. Can you provide the full unit circle values for cosine and sine?
5. How does the Pythagorean theorem apply to non-right triangles?

Tip

To verify your trigonometric calculations, always check if the square of sine and cosine adds up to 1: $\sin^2 B + \cos^2 B = 1$. This helps confirm your work!