If `alpha " and " beta ( alpha gt beta) ` are roots of the equation ` x^(2) - sqrt2 x + sqrt( 3 - 2 sqrt 2 ) = 0", then the value of " ( cos^(-1) alpha + tan^(-1) alpha + tan ^(-1) beta) ` is equal to

Options

A. `(3pi)/8`
B. `(5pi)/8`
C. `(7pi)/8`
D. `pi/3 `

Answer

To solve the problem step by step, we will find the roots of the given quadratic equation and then evaluate the expression \( \cos^{-1} \alpha + \tan^{-1} \alpha + \tan^{-1} \beta \).

### Step 1: Identify the quadratic equation
The given equation is:
\[ x^2 - \sqrt{2}x + \sqrt{3 - 2\sqrt{2}} = 0 \]

### Step 2: Simplify the constant term
We need to simplify \( \sqrt{3 - 2\sqrt{2}} \). We can rewrite it as:
\[ 3 - 2\sqrt{2} = (\sqrt{2} - 1)^2 \]
Thus,
\[ \sqrt{3 - 2\sqrt{2}} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1 \]

### Step 3: Substitute back into the equation
Now, substituting back into the quadratic equation, we get:
\[ x^2 - \sqrt{2}x + (\sqrt{2} - 1) = 0 \]

### Step 4: Apply the quadratic formula
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -\sqrt{2} \), and \( c = \sqrt{2} - 1 \).

Calculating \( b^2 - 4ac \):
\[ b^2 = (\sqrt{2})^2 = 2 \]
\[ 4ac = 4 \cdot 1 \cdot (\sqrt{2} - 1) = 4\sqrt{2} - 4 \]
Thus,
\[ b^2 - 4ac = 2 - (4\sqrt{2} - 4) = 6 - 4\sqrt{2} \]

### Step 5: Calculate the roots
Now substituting into the quadratic formula:
\[ x = \frac{\sqrt{2} \pm \sqrt{6 - 4\sqrt{2}}}{2} \]
We can simplify \( \sqrt{6 - 4\sqrt{2}} \):
\[ 6 - 4\sqrt{2} = 2(\sqrt{2} - 1)^2 \]
So,
\[ \sqrt{6 - 4\sqrt{2}} = \sqrt{2}(\sqrt{2} - 1) = 2 - \sqrt{2} \]
Thus, the roots become:
\[ x = \frac{\sqrt{2} \pm (2 - \sqrt{2})}{2} \]
Calculating the two roots:
1. \( \alpha = \frac{\sqrt{2} + (2 - \sqrt{2})}{2} = \frac{2}{2} = 1 \)
2. \( \beta = \frac{\sqrt{2} - (2 - \sqrt{2})}{2} = \frac{2\sqrt{2} - 2}{2} = \sqrt{2} - 1 \)

### Step 6: Evaluate the expression
Now we need to find:
\[ \cos^{-1} \alpha + \tan^{-1} \alpha + \tan^{-1} \beta \]
Substituting \( \alpha = 1 \) and \( \beta = \sqrt{2} - 1 \):
1. \( \cos^{-1}(1) = 0 \)
2. \( \tan^{-1}(1) = \frac{\pi}{4} \)
3. \( \tan^{-1}(\sqrt{2} - 1) = \frac{\pi}{8} \) (since \( \tan(\frac{\pi}{8}) = \sqrt{2} - 1 \))

Thus, we get:
\[ 0 + \frac{\pi}{4} + \frac{\pi}{8} = \frac{2\pi}{8} + \frac{\pi}{8} = \frac{3\pi}{8} \]

### Final Answer
The value of \( \cos^{-1} \alpha + \tan^{-1} \alpha + \tan^{-1} \beta \) is:
\[ \frac{3\pi}{8} \]