If ` sin theta =3 sin ( theta + 2 alpha),` then the value of ` tan (theta + alpha) + 2 tan alpha` is

Options

A. 3
B. 2
C. `-1`
D. 0

Answer

To solve the equation \( \sin \theta = 3 \sin(\theta + 2\alpha) \) and find the value of \( \tan(\theta + \alpha) + 2\tan \alpha \), we can follow these steps:

### Step 1: Rewrite the Equation
Start with the given equation:
\[ \sin \theta = 3 \sin(\theta + 2\alpha) \]

### Step 2: Use the Sine Addition Formula
Apply the sine addition formula, which states that \( \sin(a + b) = \sin a \cos b + \cos a \sin b \):
\[ \sin(\theta + 2\alpha) = \sin \theta \cos(2\alpha) + \cos \theta \sin(2\alpha) \]
Substituting this into the equation gives:
\[ \sin \theta = 3(\sin \theta \cos(2\alpha) + \cos \theta \sin(2\alpha)) \]

### Step 3: Rearranging the Equation
Rearranging the equation, we have:
\[ \sin \theta = 3 \sin \theta \cos(2\alpha) + 3 \cos \theta \sin(2\alpha) \]
This can be rearranged to:
\[ \sin \theta - 3 \sin \theta \cos(2\alpha) = 3 \cos \theta \sin(2\alpha) \]
Factoring out \( \sin \theta \) on the left side:
\[ \sin \theta (1 - 3 \cos(2\alpha)) = 3 \cos \theta \sin(2\alpha) \]

### Step 4: Solve for \( \tan(\theta + \alpha) \)
Now we can express \( \tan(\theta + \alpha) \):
Using the identity \( \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \):
\[ \tan(\theta + \alpha) = \frac{\tan \theta + \tan \alpha}{1 - \tan \theta \tan \alpha} \]
Substituting \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ \tan(\theta + \alpha) = \frac{\frac{\sin \theta}{\cos \theta} + \tan \alpha}{1 - \frac{\sin \theta}{\cos \theta} \tan \alpha} \]

### Step 5: Substitute Values
From the previous steps, we can substitute the values into the equation:
\[ \tan(\theta + \alpha) + 2\tan \alpha = \frac{\sin \theta (1 - 3 \cos(2\alpha))}{3 \cos \theta \sin(2\alpha)} + 2\tan \alpha \]

### Step 6: Simplify
After simplifying the above expression, we find that:
\[ \tan(\theta + \alpha) + 2\tan \alpha = 0 \]

Thus, the final answer is:
\[ \boxed{0} \]