Write all the other trigonometric ratios of `angle`B in terms of tan B.

Answer

To find all the other trigonometric ratios of angle B in terms of tan B, we can use the relationships between the trigonometric functions.

1. **Start with the definition of tan B:**
\[ \tan B = \frac{\sin B}{\cos B} \]
From this, we can express \(\sin B\) in terms of \(\tan B\) and \(\cos B\):
\[ \sin B = \tan B \cdot \cos B \]

2. **Use the Pythagorean identity:**
The Pythagorean identity states that:
\[ \sec^2 B = 1 + \tan^2 B \]
Therefore, we can express \(\sec B\) as:
\[ \sec B = \sqrt{1 + \tan^2 B} \]

3. **Find cos B:**
Since \(\sec B = \frac{1}{\cos B}\), we can find \(\cos B\):
\[ \cos B = \frac{1}{\sec B} = \frac{1}{\sqrt{1 + \tan^2 B}} \]

4. **Find sin B:**
Now, substituting \(\cos B\) back into the equation for \(\sin B\):
\[ \sin B = \tan B \cdot \cos B = \tan B \cdot \frac{1}{\sqrt{1 + \tan^2 B}} = \frac{\tan B}{\sqrt{1 + \tan^2 B}} \]

5. **Find cosec B:**
The cosecant function is the reciprocal of sine:
\[ \csc B = \frac{1}{\sin B} = \frac{\sqrt{1 + \tan^2 B}}{\tan B} \]

6. **Find sec B:**
We already found this earlier:
\[ \sec B = \sqrt{1 + \tan^2 B} \]

7. **Find cot B:**
The cotangent function is the reciprocal of tangent:
\[ \cot B = \frac{1}{\tan B} \]

Now we have all the trigonometric ratios in terms of \(\tan B\):
- \(\sin B = \frac{\tan B}{\sqrt{1 + \tan^2 B}}\)
- \(\cos B = \frac{1}{\sqrt{1 + \tan^2 B}}\)
- \(\tan B = \tan B\)
- \(\csc B = \frac{\sqrt{1 + \tan^2 B}}{\tan B}\)
- \(\sec B = \sqrt{1 + \tan^2 B}\)
- \(\cot B = \frac{1}{\tan B}\)