If `tan (alpha + theta) = n tan(alpha - theta)` , show that :`(n + 1) sin 2 theta = (n - 1) sin 2alpha`.

Answer

To solve the equation \( \tan(\alpha + \theta) = n \tan(\alpha - \theta) \) and show that \( (n + 1) \sin 2\theta = (n - 1) \sin 2\alpha \), we can follow these steps:

### Step 1: Write the given equation
We start with the given equation:
\[ \tan(\alpha + \theta) = n \tan(\alpha - \theta) \]

### Step 2: Use the tangent addition and subtraction formulas
Using the tangent addition and subtraction formulas:
\[ \tan(\alpha + \theta) = \frac{\tan \alpha + \tan \theta}{1 - \tan \alpha \tan \theta} \]
\[ \tan(\alpha - \theta) = \frac{\tan \alpha - \tan \theta}{1 + \tan \alpha \tan \theta} \]
Substituting these into the equation gives:
\[ \frac{\tan \alpha + \tan \theta}{1 - \tan \alpha \tan \theta} = n \cdot \frac{\tan \alpha - \tan \theta}{1 + \tan \alpha \tan \theta} \]

### Step 3: Cross-multiply to eliminate the fractions
Cross-multiplying yields:
\[ (\tan \alpha + \tan \theta)(1 + \tan \alpha \tan \theta) = n(\tan \alpha - \tan \theta)(1 - \tan \alpha \tan \theta) \]

### Step 4: Expand both sides
Expanding both sides results in:
\[ \tan \alpha + \tan \theta + \tan \alpha \tan^2 \theta + \tan^2 \alpha \tan \theta = n(\tan \alpha - \tan \theta - n \tan \alpha \tan^2 \theta + n \tan^2 \alpha \tan \theta) \]

### Step 5: Rearranging the equation
Rearranging terms leads to:
\[ \tan \alpha + \tan \theta + \tan \alpha \tan^2 \theta + \tan^2 \alpha \tan \theta - n(\tan \alpha - \tan \theta - n \tan \alpha \tan^2 \theta + n \tan^2 \alpha \tan \theta) = 0 \]

### Step 6: Apply the componendo and dividendo rule
Using the componendo and dividendo rule, we can simplify the equation:
\[ \frac{1 + n}{1 - n} = \frac{\tan(\alpha - \theta) + \tan(\alpha + \theta)}{\tan(\alpha - \theta) - \tan(\alpha + \theta)} \]

### Step 7: Substitute the sine and cosine
Substituting \( \tan \) in terms of sine and cosine:
\[ \frac{1 + n}{1 - n} = \frac{\frac{\sin(\alpha - \theta)}{\cos(\alpha - \theta)} + \frac{\sin(\alpha + \theta)}{\cos(\alpha + \theta)}}{\frac{\sin(\alpha - \theta)}{\cos(\alpha - \theta)} - \frac{\sin(\alpha + \theta)}{\cos(\alpha + \theta)}} \]

### Step 8: Simplify the right-hand side
This can be simplified to:
\[ \frac{(1 + n) \sin 2\theta}{(1 - n) \sin 2\alpha} \]

### Step 9: Final rearrangement
Finally, we rearrange to show:
\[ (n + 1) \sin 2\theta = (n - 1) \sin 2\alpha \]

### Conclusion
Thus, we have shown that:
\[ (n + 1) \sin 2\theta = (n - 1) \sin 2\alpha \]
Hence proved.

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