Show that (SinA+sin 2A)/1+cosA+cos2A =tanA

We are tasked with proving the following identity:

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

### Step-by-step proof:

#### Step 1: Apply the double angle formulas for sine and cosine.

We know the double angle identities:
\[
\sin 2A = 2 \sin A \cos A
\]
\[
\cos 2A = 2 \cos^2 A - 1
\]

Substitute these into the equation:

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

#### Step 2: Simplify the denominator.
\[
1 + \cos A + 2 \cos^2 A - 1 = \cos A + 2 \cos^2 A
\]

So, the expression becomes:
\[
\frac{\sin A (1 + 2 \cos A)}{\cos A (1 + 2 \cos A)}
\]

#### Step 3: Cancel common factors.

Since \(1 + 2 \cos A\) is common in both the numerator and the denominator, we can cancel it:

\[
\frac{\sin A}{\cos A}
\]

#### Step 4: Recognize the tangent identity.

We know that:
\[
\frac{\sin A}{\cos A} = \tan A
\]

Thus, we have proven that:
\[
\frac{\sin A + \sin 2A}{1 + \cos A + \cos 2A} = \tan A
\]

### Conclusion:
The identity holds true.

Would you like more details or further explanation?

Here are 5 related questions you might find useful:
1. How are double angle formulas derived?
2. Can you explain more about tangent identities?
3. How would the identity change if \( A \) were replaced by \( B \)?
4. What happens if we apply this identity to specific angle values?
5. How do we derive other trigonometric identities using similar methods?

**Tip:** When proving trigonometric identities, substituting known identities for terms like \(\sin 2A\) or \(\cos 2A\) is often a key step!

Show that (SinA+sin 2A)/1+cosA+cos2A = tanA

In this detailed step-by-step proof, learn how to demonstrate the trigonometric identity (SinA + sin 2A) / (1 + cosA + cos 2A) = tanA using double angle formulas and tangent identities. Ideal for students in Grades 10-12.

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