If `alpha` and `beta` are the roots of equation `(k+1) tan^(2)x-sqrt2lambda tan(x)=1-k` and `tan^(2)(alpha+beta)=50`. Find the value of `lambda`

Options

A. `10sqrt(2)`
B. `5sqrt(2)`
C. `10`
D. 5

Answer

To solve the problem step by step, we will analyze the given quadratic equation and use the properties of the roots.

### Step 1: Write the given quadratic equation
The equation given is:
\[ (k + 1) \tan^2 x - \sqrt{2} \lambda \tan x = 1 - k \]
Rearranging this, we get:
\[ (k + 1) \tan^2 x - \sqrt{2} \lambda \tan x + (k - 1) = 0 \]

### Step 2: Identify the coefficients
From the standard form of a quadratic equation \( ax^2 + bx + c = 0 \), we identify:
- \( a = k + 1 \)
- \( b = -\sqrt{2} \lambda \)
- \( c = k - 1 \)

### Step 3: Use the properties of roots
For a quadratic equation, the sum and product of the roots can be expressed as:
- Sum of roots \( \tan \alpha + \tan \beta = -\frac{b}{a} = \frac{\sqrt{2} \lambda}{k + 1} \)
- Product of roots \( \tan \alpha \tan \beta = \frac{c}{a} = \frac{k - 1}{k + 1} \)

### Step 4: Find \( \tan(\alpha + \beta) \)
Using the formula for \( \tan(\alpha + \beta) \):
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
Substituting the values we found:
\[ \tan(\alpha + \beta) = \frac{\frac{\sqrt{2} \lambda}{k + 1}}{1 - \frac{k - 1}{k + 1}} \]

### Step 5: Simplify the expression
The denominator simplifies as follows:
\[ 1 - \frac{k - 1}{k + 1} = \frac{(k + 1) - (k - 1)}{k + 1} = \frac{2}{k + 1} \]
Thus, we have:
\[ \tan(\alpha + \beta) = \frac{\frac{\sqrt{2} \lambda}{k + 1}}{\frac{2}{k + 1}} = \frac{\sqrt{2} \lambda}{2} \]

### Step 6: Find \( \tan^2(\alpha + \beta) \)
Now, squaring both sides gives:
\[ \tan^2(\alpha + \beta) = \left(\frac{\sqrt{2} \lambda}{2}\right)^2 = \frac{2 \lambda^2}{4} = \frac{\lambda^2}{2} \]

### Step 7: Set the equation with the given value
We are given that:
\[ \tan^2(\alpha + \beta) = 50 \]
Thus, we equate:
\[ \frac{\lambda^2}{2} = 50 \]

### Step 8: Solve for \( \lambda \)
Multiplying both sides by 2:
\[ \lambda^2 = 100 \]
Taking the square root:
\[ \lambda = 10 \]

### Final Answer
The value of \( \lambda \) is:
\[ \boxed{10} \]