Find the value(s) of 'a' for which the inequality `tan^(2)x+(a+1)tan x - (a-3)lt 0`, is true for at least on `x in (0, pi//2)`.

Answer

To solve the inequality \( \tan^2 x + (a + 1) \tan x - (a - 3) < 0 \) for at least one \( x \in (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Substitute \( \tan x \) with \( t \)
Let \( t = \tan x \). As \( x \) varies from \( 0 \) to \( \frac{\pi}{2} \), \( t \) varies from \( 0 \) to \( \infty \). The inequality now becomes:
\[ t^2 + (a + 1)t - (a - 3) < 0 \]

### Step 2: Analyze the quadratic inequality
The quadratic inequality \( t^2 + (a + 1)t - (a - 3) < 0 \) is a parabola that opens upwards (since the coefficient of \( t^2 \) is positive). For this inequality to hold for at least one \( t \) in \( (0, \infty) \), the quadratic must have real roots and the value of the quadratic must be negative between the roots.

### Step 3: Find the discriminant
The discriminant \( D \) of the quadratic equation \( at^2 + bt + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = a + 1 \), and \( c = -(a - 3) \). Therefore, the discriminant is:
\[ D = (a + 1)^2 - 4 \cdot 1 \cdot (-(a - 3)) \]
\[ D = (a + 1)^2 + 4(a - 3) \]
\[ D = a^2 + 2a + 1 + 4a - 12 \]
\[ D = a^2 + 6a - 11 \]

### Step 4: Set the discriminant greater than zero
For the quadratic to have real roots, we need:
\[ a^2 + 6a - 11 > 0 \]
We will solve this inequality by first finding the roots of the equation \( a^2 + 6a - 11 = 0 \) using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{(6)^2 - 4 \cdot 1 \cdot (-11)}}{2 \cdot 1} \]
\[ = \frac{-6 \pm \sqrt{36 + 44}}{2} = \frac{-6 \pm \sqrt{80}}{2} = \frac{-6 \pm 4\sqrt{5}}{2} \]
\[ = -3 \pm 2\sqrt{5} \]
Thus, the roots are:
\[ a_1 = -3 - 2\sqrt{5}, \quad a_2 = -3 + 2\sqrt{5} \]

### Step 5: Analyze the intervals
The quadratic \( a^2 + 6a - 11 \) opens upwards, so it is positive outside the roots:
\[ a < -3 - 2\sqrt{5} \quad \text{or} \quad a > -3 + 2\sqrt{5} \]

### Step 6: Check the value of the quadratic at \( t = 0 \)
Next, we check the value of the quadratic at \( t = 0 \):
\[ f(0) = -(a - 3) < 0 \implies a - 3 > 0 \implies a > 3 \]

### Step 7: Combine the conditions
Now we combine the conditions:
1. \( a < -3 - 2\sqrt{5} \) or \( a > -3 + 2\sqrt{5} \)
2. \( a > 3 \)

Since \( -3 + 2\sqrt{5} \) is approximately \( -3 + 4.472 = 1.472 \), we find that:
- The condition \( a > 3 \) is stronger than \( a > -3 + 2\sqrt{5} \).

### Final Solution
Thus, the final solution for \( a \) is:
\[ a > 3 \]