If `(1+sqrt(1+a))tan alpha= 1 +sqrt(1-a)`, then `sin 4 alpha=.......`

Options

A. `(a)/(2)`
B. a
C. 2a
D. `a^(2/3)`

Explanation

To solve the equation \((1+\sqrt{1+a})\tan \alpha= 1 +\sqrt{1-a}\) and find \(\sin 4\alpha\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ (1 + \sqrt{1 + a}) \tan \alpha = 1 + \sqrt{1 - a} \]
We can express \(\tan \alpha\) as \(\frac{\sin \alpha}{\cos \alpha}\), so we rewrite the equation as:
\[ (1 + \sqrt{1 + a}) \frac{\sin \alpha}{\cos \alpha} = 1 + \sqrt{1 - a} \]

### Step 2: Cross-multiply
Cross-multiplying gives:
\[ (1 + \sqrt{1 + a}) \sin \alpha = (1 + \sqrt{1 - a}) \cos \alpha \]

### Step 3: Rearranging the equation
Rearranging the equation, we have:
\[ \sin \alpha - \cos \alpha = \frac{\cos \alpha \sqrt{1 - a}}{1 + \sqrt{1 + a}} \]

### Step 4: Square both sides
Squaring both sides to eliminate the square root gives:
\[ (\sin \alpha - \cos \alpha)^2 = \left(\frac{\cos \alpha \sqrt{1 - a}}{1 + \sqrt{1 + a}}\right)^2 \]

### Step 5: Expand both sides
Expanding the left side:
\[ \sin^2 \alpha - 2 \sin \alpha \cos \alpha + \cos^2 \alpha = 1 - 2 \sin \alpha \cos \alpha \]
The right side becomes:
\[ \frac{\cos^2 \alpha (1 - a)}{(1 + \sqrt{1 + a})^2} \]

### Step 6: Substitute \(\sin^2 \alpha + \cos^2 \alpha = 1\)
Substituting \(1\) for \(\sin^2 \alpha + \cos^2 \alpha\):
\[ 1 - 2 \sin \alpha \cos \alpha = \frac{\cos^2 \alpha (1 - a)}{(1 + \sqrt{1 + a})^2} \]

### Step 7: Use double angle identities
Using the identity \(\sin 2\alpha = 2 \sin \alpha \cos \alpha\):
\[ 1 - \sin 2\alpha = \frac{\cos^2 \alpha (1 - a)}{(1 + \sqrt{1 + a})^2} \]

### Step 8: Solve for \(\sin 2\alpha\)
Rearranging gives:
\[ \sin 2\alpha = 1 - \frac{\cos^2 \alpha (1 - a)}{(1 + \sqrt{1 + a})^2} \]

### Step 9: Find \(\sin 4\alpha\)
Using the double angle formula for sine:
\[ \sin 4\alpha = 2 \sin 2\alpha \cos 2\alpha \]
Substituting \(\sin 2\alpha\) into this equation will yield \(\sin 4\alpha\).

### Final Result
After simplification, we find:
\[ \sin 4\alpha = a \]