If `e^({(sin^2 x + sin^4x + sin^6x+...oo)log_e2})`satisfies the equation `x^2 - 9x + 8 = 0` , then value of
`(cosx)/(cosx+sinx ), 0 le x le pi/2` is

Options

A. `1/2 (sqrt3+1)`
B. `1/2(sqrt3 -1)`
C. `1/2`
D. 0

Answer

To solve the given problem step by step, we will follow the reasoning outlined in the video transcript and derive the solution systematically.

### Step 1: Calculate the Infinite Series
We start with the infinite series given by:
\[ \sin^2 x + \sin^4 x + \sin^6 x + \ldots \]
This series can be recognized as a geometric series where:
- The first term \( a = \sin^2 x \)
- The common ratio \( r = \sin^2 x \)

The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{\sin^2 x}{1 - \sin^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x \]

### Step 2: Substitute into the Exponential Expression
Now, we substitute this result into the expression:
\[ e^{(\sin^2 x + \sin^4 x + \sin^6 x + \ldots) \log_e 2} = e^{\tan^2 x \log_e 2} \]
Using the property of exponents, we can simplify this to:
\[ = 2^{\tan^2 x} \]

### Step 3: Set Up the Quadratic Equation
We know that this expression satisfies the quadratic equation:
\[ x^2 - 9x + 8 = 0 \]
To find the roots of this equation, we can factor it:
\[ (x - 8)(x - 1) = 0 \]
Thus, the roots are:
\[ x = 8 \quad \text{and} \quad x = 1 \]

### Step 4: Solve for \( \tan^2 x \)
Now, we set the expression \( 2^{\tan^2 x} \) equal to the roots:
1. For \( 2^{\tan^2 x} = 8 \):
\[ \tan^2 x = 3 \quad \Rightarrow \quad \tan x = \sqrt{3} \quad \Rightarrow \quad x = \frac{\pi}{3} \]
2. For \( 2^{\tan^2 x} = 1 \):
\[ \tan^2 x = 0 \quad \Rightarrow \quad \tan x = 0 \quad \Rightarrow \quad x = 0 \]

### Step 5: Calculate \( \frac{\cos x}{\cos x + \sin x} \)
Next, we need to evaluate:
\[ \frac{\cos x}{\cos x + \sin x} \]

#### Case 1: When \( x = \frac{\pi}{3} \)
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \frac{\cos\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right) + \sin\left(\frac{\pi}{3}\right)} = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{\sqrt{3}}{2}} = \frac{\frac{1}{2}}{\frac{1 + \sqrt{3}}{2}} = \frac{1}{1 + \sqrt{3}} \]
To rationalize the denominator:
\[ \frac{1}{1 + \sqrt{3}} \cdot \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{1 - \sqrt{3}}{1 - 3} = \frac{1 - \sqrt{3}}{-2} = \frac{\sqrt{3} - 1}{2} \]

#### Case 2: When \( x = 0 \)
\[ \cos(0) = 1, \quad \sin(0) = 0 \]
Thus,
\[ \frac{\cos(0)}{\cos(0) + \sin(0)} = \frac{1}{1 + 0} = 1 \]

### Final Result
The values we found are:
- For \( x = \frac{\pi}{3} \): \( \frac{\sqrt{3} - 1}{2} \)
- For \( x = 0 \): \( 1 \)

Since both values are valid within the interval \( [0, \frac{\pi}{2}] \), the final answer can be expressed as:
\[ \text{The value of } \frac{\cos x}{\cos x + \sin x} \text{ is } \frac{\sqrt{3} - 1}{2} \text{ or } 1. \]