`(sintheta+sin2theta)/(1+costheta+cos2theta)=?`

Options

A. `tantheta`
B. `sintheta`
C. `costheta`
D. `tan^(2)theta`

Answer

To solve the expression \(\frac{\sin \theta + \sin 2\theta}{1 + \cos \theta + \cos 2\theta}\), we will follow these steps:

### Step 1: Substitute \(\sin 2\theta\) and \(\cos 2\theta\)
We know that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
\[ \cos 2\theta = 2 \cos^2 \theta - 1 \]
Substituting these into the expression gives:
\[ \frac{\sin \theta + 2 \sin \theta \cos \theta}{1 + \cos \theta + (2 \cos^2 \theta - 1)} \]

### Step 2: Simplify the denominator
The denominator simplifies as follows:
\[ 1 + \cos \theta + 2 \cos^2 \theta - 1 = \cos \theta + 2 \cos^2 \theta \]
So now our expression looks like:
\[ \frac{\sin \theta (1 + 2 \cos \theta)}{\cos \theta + 2 \cos^2 \theta} \]

### Step 3: Factor out common terms
In the denominator, we can factor out \(\cos \theta\):
\[ \cos \theta + 2 \cos^2 \theta = \cos \theta(1 + 2 \cos \theta) \]
Thus, our expression becomes:
\[ \frac{\sin \theta (1 + 2 \cos \theta)}{\cos \theta (1 + 2 \cos \theta)} \]

### Step 4: Cancel common terms
We can cancel \(1 + 2 \cos \theta\) from the numerator and the denominator (assuming \(1 + 2 \cos \theta \neq 0\)):
\[ \frac{\sin \theta}{\cos \theta} \]

### Step 5: Final result
This simplifies to:
\[ \tan \theta \]

Thus, the final answer is:
\[ \frac{\sin \theta + \sin 2\theta}{1 + \cos \theta + \cos 2\theta} = \tan \theta \]

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