lf α and β are two different solutions lying between `−π/2` and `π/2 ` of the equation ` 2tanθ+Secθ=2 ` then ` tanα+tanβ `

Answer

To solve the equation \( 2\tan\theta + \sec\theta = 2 \) and find \( \tan\alpha + \tan\beta \) where \( \alpha \) and \( \beta \) are the two different solutions lying between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), we can follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ 2\tan\theta + \sec\theta = 2 \]
Rearranging gives:
\[ \sec\theta = 2 - 2\tan\theta \]

### Step 2: Square both sides
Next, we square both sides to eliminate the secant:
\[ \sec^2\theta = (2 - 2\tan\theta)^2 \]
Expanding the right side:
\[ \sec^2\theta = 4 - 8\tan\theta + 4\tan^2\theta \]

### Step 3: Use the identity for secant
Recall the identity \( \sec^2\theta = 1 + \tan^2\theta \). Substitute this into the equation:
\[ 1 + \tan^2\theta = 4 - 8\tan\theta + 4\tan^2\theta \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 0 = 4\tan^2\theta - 8\tan\theta + 3 \]
This can be rewritten as:
\[ 4\tan^2\theta - 8\tan\theta + 3 = 0 \]

### Step 5: Apply the quadratic formula
Now, we can apply the quadratic formula \( \tan\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4 \), \( b = -8 \), and \( c = 3 \):
\[ \tan\theta = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4} \]
Calculating the discriminant:
\[ \tan\theta = \frac{8 \pm \sqrt{64 - 48}}{8} = \frac{8 \pm \sqrt{16}}{8} = \frac{8 \pm 4}{8} \]

### Step 6: Find the solutions
This gives us two solutions:
\[ \tan\theta_1 = \frac{12}{8} = \frac{3}{2}, \quad \tan\theta_2 = \frac{4}{8} = \frac{1}{2} \]

### Step 7: Calculate \( \tan\alpha + \tan\beta \)
Let \( \tan\alpha = \frac{3}{2} \) and \( \tan\beta = \frac{1}{2} \). Thus:
\[ \tan\alpha + \tan\beta = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2 \]

### Final Answer
Therefore, the value of \( \tan\alpha + \tan\beta \) is:
\[ \boxed{2} \]