If exp `[(sin^(2)x+sin^(4) x +sin^(6)x+.... oo)" In" 2]` satisfies the equation `y^(2)-9y+8=0`, then the value of `(cos x )/( cos x+ sin x ),0 lt x lt (pi)/(2)` , is

Options

A. `sqrt(3)+1 `
B. `(sqrt(3)-1)/(2)`
C. `sqrt(3)-1 `
D. None of these

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript and break it down into clear steps:

### Step 1: Identify the Infinite Series
We start with the expression:
\[ \exp\left(\sum_{n=1}^{\infty} \sin^{2n} x\right) \ln 2 \]
The series \( \sin^2 x + \sin^4 x + \sin^6 x + \ldots \) is a geometric series where the first term \( a = \sin^2 x \) and the common ratio \( r = \sin^2 x \).

### Step 2: Sum of the Infinite Series
The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Thus, we have:
\[ \sum_{n=1}^{\infty} \sin^{2n} x = \frac{\sin^2 x}{1 - \sin^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x \]

### Step 3: Substitute Back into the Exponential
Now substituting back into the exponential:
\[ \exp\left(\tan^2 x\right) \ln 2 \]

### Step 4: Let \( y = 2^{\tan^2 x} \)
We can rewrite the expression as:
\[ y = 2^{\tan^2 x} \]

### Step 5: Solve the Quadratic Equation
We know that \( y \) satisfies the equation:
\[ y^2 - 9y + 8 = 0 \]
To find the roots of this quadratic equation, we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -9, c = 8 \):
\[ y = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{9 \pm \sqrt{81 - 32}}{2} = \frac{9 \pm \sqrt{49}}{2} = \frac{9 \pm 7}{2} \]
Thus, the roots are:
\[ y = 8 \quad \text{and} \quad y = 1 \]

### Step 6: Find \( \tan^2 x \) for Each Root
1. For \( y = 1 \):
\[ 1 = 2^{\tan^2 x} \implies \tan^2 x = 0 \implies x = 0 \]
(Not valid since \( 0 < x < \frac{\pi}{2} \))

2. For \( y = 8 \):
\[ 8 = 2^{\tan^2 x} \implies \tan^2 x = 3 \implies \tan x = \sqrt{3} \]
Thus, \( x = \frac{\pi}{3} \).

### Step 7: Calculate the Desired Value
We need to find:
\[ \frac{\cos x}{\cos x + \sin x} \]
Substituting \( x = \frac{\pi}{3} \):
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Thus:
\[ \frac{\cos\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right) + \sin\left(\frac{\pi}{3}\right)} = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{\sqrt{3}}{2}} = \frac{\frac{1}{2}}{\frac{1 + \sqrt{3}}{2}} = \frac{1}{1 + \sqrt{3}} \]

### Final Answer
The value of \( \frac{\cos x}{\cos x + \sin x} \) is:
\[ \frac{1}{1 + \sqrt{3}} \]