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The given problem provides:

\[
\sin A = \frac{1}{8}, \quad 0^\circ < A < 90^\circ
\]

We are asked to find \(\tan \frac{A}{2}\).

### Solution:
We can use the half-angle identity for tangent:

\[
\tan \frac{A}{2} = \frac{1 - \cos A}{\sin A}
\]

#### Step 1: Find \(\cos A\)

We use the Pythagorean identity for sine and cosine:

\[
\sin^2 A + \cos^2 A = 1
\]

Substituting \(\sin A = \frac{1}{8}\):

\[
\left(\frac{1}{8}\right)^2 + \cos^2 A = 1
\]

\[
\frac{1}{64} + \cos^2 A = 1
\]

\[
\cos^2 A = 1 - \frac{1}{64} = \frac{63}{64}
\]

\[
\cos A = \sqrt{\frac{63}{64}} = \frac{\sqrt{63}}{8}
\]

#### Step 2: Apply the half-angle formula

Now, substitute \(\sin A = \frac{1}{8}\) and \(\cos A = \frac{\sqrt{63}}{8}\) into the half-angle identity:

\[
\tan \frac{A}{2} = \frac{1 - \frac{\sqrt{63}}{8}}{\frac{1}{8}}
\]

Simplifying the expression:

\[
\tan \frac{A}{2} = \frac{8 - \sqrt{63}}{1}
\]

\[
\tan \frac{A}{2} = 8 - \sqrt{63}
\]

Thus, the value of \(\tan \frac{A}{2}\) is:

\[
\tan \frac{A}{2} = 8 - \sqrt{63}
\]

Would you like more details or have any questions?

Here are some related questions:

1. How can you derive the half-angle identities for sine and cosine?
2. How does the Pythagorean identity help in trigonometric calculations?
3. What are some other applications of the half-angle identities in trigonometry?
4. How can we find the angle \( A \) if we only know \(\sin A\)?
5. How do other trigonometric functions change when using half-angle formulas?

**Tip**: When solving trigonometric identities, always ensure that the angle falls within the proper quadrant to avoid sign errors in sine, cosine, or tangent values.

Given: sin A = 1/8, 0° < A < 90°. Find: tan(A/2).

Solve a trigonometry problem where sin(A) = 1/8 and 0° < A < 90°. Learn to find tan(A/2) using the half-angle identity and Pythagorean identity. Suitable for high school students in Grades 10-12.

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