Given a right triangle $ABC$ with a right angle at $A$, and $\cos C = \frac{1}{3}$, find the values of $\sin B$, $\tan B$, and $\cot B$.

1. Problem Description

Given a right triangle $ABC$ with a right angle at $A$, and $\cos C = \frac{1}{3}$, find the values of $\sin B$, $\tan B$, and $\cot B$.

2. Solution Steps

Since $ABC$ is a right triangle with a right angle at $A$, we have $A + B + C = 180^\circ$, and $A = 90^\circ$, so $B + C = 90^\circ$. This means $B$ and $C$ are complementary angles.

Therefore, $\sin B = \cos C$. Since $\cos C = \frac{1}{3}$, we have $\sin B = \frac{1}{3}$.

We want to find $\tan B$ and $\cot B$. We know that $\cot B = \frac{1}{\tan B}$.

Also, $\tan B = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{AB}$.

Since $\cos C = \frac{1}{3}$, we can assume $AC = 1$ and $BC = 3$.

Using the Pythagorean theorem, $AB^2 + AC^2 = BC^2$, so $AB^2 + 1^2 = 3^2$.

Then $AB^2 = 9 - 1 = 8$, so $AB = \sqrt{8} = 2\sqrt{2}$.

Now we can find $\tan B = \frac{AC}{AB} = \frac{1}{2\sqrt{2}} = \frac{1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4}$.

Then $\cot B = \frac{1}{\tan B} = \frac{1}{\frac{\sqrt{2}}{4}} = \frac{4}{\sqrt{2}} = \frac{4}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.

We can also solve for $\tan B$ and $\cot B$ by using the relation $B+C = 90^{\circ}$.

Then $\tan B = \tan(90^{\circ} - C) = \cot C$.

Since $\cos C = \frac{1}{3}$, we have $\cos^2 C = \frac{1}{9}$.

Also $\sin^2 C + \cos^2 C = 1$, so $\sin^2 C = 1 - \frac{1}{9} = \frac{8}{9}$.

Therefore $\sin C = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.

Then $\tan C = \frac{\sin C}{\cos C} = \frac{\frac{2\sqrt{2}}{3}}{\frac{1}{3}} = 2\sqrt{2}$.

So $\cot C = \frac{1}{\tan C} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$.

We have $\tan B = \cot C = 2\sqrt{2}$ and $\cot B = \tan C = \frac{\sqrt{2}}{4}$.

We made an error somewhere.

Since $\sin B = \cos C = \frac{1}{3}$, we can compute $\sin^2 B = \frac{1}{9}$.

Then $\cos^2 B = 1 - \sin^2 B = 1 - \frac{1}{9} = \frac{8}{9}$.

Then $\cos B = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.

So $\tan B = \frac{\sin B}{\cos B} = \frac{\frac{1}{3}}{\frac{2\sqrt{2}}{3}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$.

$\cot B = \frac{1}{\tan B} = 2\sqrt{2}$.

3. Final Answer

$\sin B = \frac{1}{3}$

$\tan B = \frac{\sqrt{2}}{4}$

$\cot B = 2\sqrt{2}$