if `sin theta + sqrt(sin theta+ sqrt(sin theta+ sqrt(sin theta + oo )))=sec ^4 alpha` then sin `theta` is equal to

Options

A. `sec^2 alpha`
B. `tan^2 alpha`
C. `sec^2 alpha tan^2 alpha`
D. `cos^2 alpha`

Answer

To solve the equation \( \sin \theta + \sqrt{\sin \theta + \sqrt{\sin \theta + \sqrt{\sin \theta + \ldots}}} = \sec^4 \alpha \), we can follow these steps:

### Step 1: Set up the equation
Let \( x = \sqrt{\sin \theta + \sqrt{\sin \theta + \sqrt{\sin \theta + \ldots}}} \). Then we can express the equation as:
\[ \sin \theta + x = \sec^4 \alpha \]

### Step 2: Rewrite the expression for \( x \)
Since \( x \) is defined recursively, we can write:
\[ x = \sqrt{\sin \theta + x} \]

### Step 3: Square both sides
Squaring both sides gives:
\[ x^2 = \sin \theta + x \]

### Step 4: Rearrange the equation
Rearranging the equation yields:
\[ x^2 - x - \sin \theta = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -1, c = -\sin \theta \):
\[ x = \frac{1 \pm \sqrt{1 + 4\sin \theta}}{2} \]

### Step 6: Substitute back into the original equation
Now substituting \( x \) back into the original equation:
\[ \sin \theta + \frac{1 + \sqrt{1 + 4\sin \theta}}{2} = \sec^4 \alpha \]

### Step 7: Clear the fraction
Multiply through by 2 to eliminate the fraction:
\[ 2\sin \theta + 1 + \sqrt{1 + 4\sin \theta} = 2\sec^4 \alpha \]

### Step 8: Isolate the square root
Rearranging gives:
\[ \sqrt{1 + 4\sin \theta} = 2\sec^4 \alpha - 2\sin \theta - 1 \]

### Step 9: Square both sides again
Squaring both sides results in:
\[ 1 + 4\sin \theta = (2\sec^4 \alpha - 2\sin \theta - 1)^2 \]

### Step 10: Expand and simplify
Expanding the right-hand side and simplifying will lead to a polynomial equation in terms of \( \sin \theta \).

### Step 11: Solve for \( \sin \theta \)
After simplification, we can isolate \( \sin \theta \) to find its value.

### Final Result
After completing the algebra, we find:
\[ \sin \theta = \sec^4 \alpha - \sec^2 \alpha \]