To find the value of \(\tan B\) when \(\tan A = 1\), we will use the expression for \(\tan B\) in terms of \(\tan A\) derived earlier:

\[
\tan B = \frac{1 - \tan A}{\tan A^2 + \tan A}
\]

Given \(\tan A = 1\), substitute this into the formula:

\[
\tan B = \frac{1 - 1}{1^2 + 1}
\]

Simplify the expression:

\[
\tan B = \frac{0}{1 + 1} = \frac{0}{2} = 0
\]

Therefore, the value of \(\tan B\) is:

\[
\tan B = 0
\]

Would you like more details or have any questions about this solution?

Here are 5 related questions to expand your understanding:
1. What does \(\tan B = 0\) imply about the angle \(B\)?
2. How does the value of \(\tan B\) change if \(\tan A = -1\)?
3. What happens to \(\tan B\) when \(\tan A = 0\)?
4. How can the values of \(\tan A\) and \(\tan B\) determine the quadrant in which angles \(A\) and \(B\) are located?
5. Can you find an expression for \(\cot B\) in terms of \(\tan A\)?

**Tip:** When solving trigonometric problems, keep in mind the behavior of trigonometric functions within different quadrants to better understand their values and relationships.

Discover how to find the value of \(\tan B\) when given that \(\tan A = 1\). This trigonometry problem involves using the tangent function and understanding its properties. Suitable for students in Grades 10-12.

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