Solve (sinA+sin2A)/(1+cosA+cos2A) is the equivalent to tanA 

We are given the expression:

\[
\frac{\sin A + \sin 2A}{1 + \cos A + \cos 2A}
\]

We need to prove that this expression is equivalent to \(\tan A\).

### Step-by-step solution:

1. **Use the double-angle identities**:
   - \(\sin 2A = 2\sin A \cos A\)
   - \(\cos 2A = 2\cos^2 A - 1\)

Substituting these into the original expression:

\[
\frac{\sin A + 2\sin A \cos A}{1 + \cos A + (2\cos^2 A - 1)}
\]

2. **Simplify the denominator**:
   The denominator becomes:
   \[
   1 + \cos A + 2\cos^2 A - 1 = \cos A + 2\cos^2 A
   \]

So the expression is now:

\[
\frac{\sin A(1 + 2\cos A)}{\cos A(1 + 2\cos A)}
\]

3. **Cancel out the common factor \((1 + 2\cos A)\)** from both the numerator and the denominator (assuming it is non-zero):
   \[
   \frac{\sin A}{\cos A}
   \]

4. **Final step**:
   \(\frac{\sin A}{\cos A} = \tan A\)

Thus, we have proved that:

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

### Conclusion:
The given expression is equivalent to \(\tan A\).

Would you like more details or further clarification?

Here are some related questions:
1. What are the other double-angle identities for sine and cosine?
2. How do you derive the identity for \(\tan(A + B)\)?
3. Can this method be applied to other trigonometric identities?
4. What happens if \(1 + 2\cos A = 0\)?
5. What is the geometric interpretation of \(\tan A\) on the unit circle?

**Tip**: Always remember that double-angle identities are useful tools in simplifying trigonometric expressions.

Solve (sinA+sin2A)/(1+cosA+cos2A) is the equivalent to tanA.

Learn how to solve the trigonometric identity (sinA + sin2A) / (1 + cosA + cos2A) and prove that it is equivalent to tanA. This solution covers double-angle identities and the simplification of trigonometric expressions, suitable for high school students in Grades 10-12.

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