Let `alpha` and `beta` be the roots of the quadratic equation `x^(2)` sin `theta - x (sin theta cos theta + 1) + cos theta = 0 (0 lt theta lt 45^(@))`, and `alpha lt beta`.
Then `Sigma_(n=0)^(oo) (alpha^(n) + ((-1)^(n))/(beta^(n)))` is equal to

Options

A. `(1)/(1 - cos theta) + (1)/(1 + sin theta)`
B. `(1)/(1 + cos theta) + (1)/(1 - sin theta)`
C. `(1)/(1 - cos theta) - (1)/(1 + sin theta)`
D. `(1)/(1 + cos theta) - (1)/(1 - sin theta)`

Answer

To solve the given problem, we start with the quadratic equation:

\[ x^2 \sin \theta - x (\sin \theta \cos \theta + 1) + \cos \theta = 0 \]

### Step 1: Identify the roots of the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we identify \( a = \sin \theta \), \( b = -(\sin \theta \cos \theta + 1) \), and \( c = \cos \theta \).

Calculating the discriminant:

\[ D = b^2 - 4ac = [-(\sin \theta \cos \theta + 1)]^2 - 4(\sin \theta)(\cos \theta) \]

Calculating \( b^2 \):

\[ b^2 = (\sin \theta \cos \theta + 1)^2 = \sin^2 \theta \cos^2 \theta + 2\sin \theta \cos \theta + 1 \]

Calculating \( 4ac \):

\[ 4ac = 4\sin \theta \cos \theta \]

Thus, the discriminant becomes:

\[ D = \sin^2 \theta \cos^2 \theta + 2\sin \theta \cos \theta + 1 - 4\sin \theta \cos \theta \]
\[ D = \sin^2 \theta \cos^2 \theta - 2\sin \theta \cos \theta + 1 \]
\[ D = (\sin \theta \cos \theta - 1)^2 \]

### Step 2: Calculate the roots
Now, substituting back into the quadratic formula:

\[ x = \frac{(\sin \theta \cos \theta + 1) \pm (\sin \theta \cos \theta - 1)}{2\sin \theta} \]

Calculating the two roots:

1. For the positive root:
\[ x_1 = \frac{2\sin \theta \cos \theta}{2\sin \theta} = \cos \theta \]

2. For the negative root:
\[ x_2 = \frac{2}{2\sin \theta} = \frac{1}{\sin \theta} = \csc \theta \]

Thus, we have:
\[ \alpha = \cos \theta, \quad \beta = \csc \theta \]

### Step 3: Find the summation
We need to evaluate the series:

\[ \sum_{n=0}^{\infty} \left( \alpha^n + \frac{(-1)^n}{\beta^n} \right) \]

This can be separated into two series:

1. The first series:
\[ \sum_{n=0}^{\infty} \alpha^n = \frac{1}{1 - \alpha} = \frac{1}{1 - \cos \theta} \]

2. The second series:
\[ \sum_{n=0}^{\infty} \frac{(-1)^n}{\beta^n} = \frac{1}{1 + \frac{1}{\beta}} = \frac{\beta}{\beta + 1} = \frac{\csc \theta}{\csc \theta + 1} \]

### Step 4: Combine the results
Combining both series gives:

\[ \sum_{n=0}^{\infty} \left( \alpha^n + \frac{(-1)^n}{\beta^n} \right) = \frac{1}{1 - \cos \theta} + \frac{\csc \theta}{\csc \theta + 1} \]

### Step 5: Simplify the expression
Using \( \csc \theta = \frac{1}{\sin \theta} \):

\[ \frac{\csc \theta}{\csc \theta + 1} = \frac{\frac{1}{\sin \theta}}{\frac{1}{\sin \theta} + 1} = \frac{1}{1 + \sin \theta} \]

Thus, the final result is:

\[ \sum_{n=0}^{\infty} \left( \alpha^n + \frac{(-1)^n}{\beta^n} \right) = \frac{1}{1 - \cos \theta} + \frac{1}{1 + \sin \theta} \]