If x satisfies the inequality `(tan^-1x)^2+3(tan^-1x)-4gt0` , then the complete set of values of x is

Options

A. `(-tan4,pi/4)`
B. `(oo,tan4)cup(pi/4,oo)`
C. `(tan1,oo)`
D. `(tan4,tan1)`

Answer

To solve the inequality \((\tan^{-1}x)^2 + 3(\tan^{-1}x) - 4 > 0\), we can follow these steps:

### Step 1: Substitute \(y\) for \(\tan^{-1}x\)
Let \(y = \tan^{-1}x\). Then the inequality becomes:
\[ y^2 + 3y - 4 > 0 \]

### Step 2: Factor the quadratic expression
We need to factor the quadratic expression \(y^2 + 3y - 4\). We can do this by finding two numbers that multiply to \(-4\) (the constant term) and add to \(3\) (the coefficient of \(y\)). The numbers \(4\) and \(-1\) satisfy this condition. Therefore, we can rewrite the inequality as:
\[ (y - 1)(y + 4) > 0 \]

### Step 3: Identify critical points
The critical points of the inequality are found by setting each factor to zero:
1. \(y - 1 = 0 \Rightarrow y = 1\)
2. \(y + 4 = 0 \Rightarrow y = -4\)

### Step 4: Test intervals
We will test the intervals determined by the critical points \(-4\) and \(1\):
- Interval 1: \(y < -4\) (e.g., \(y = -5\))
- Interval 2: \(-4 < y < 1\) (e.g., \(y = 0\))
- Interval 3: \(y > 1\) (e.g., \(y = 2\))

**Testing Interval 1:**
For \(y = -5\):
\[ (-5 - 1)(-5 + 4) = (-6)(-1) = 6 > 0 \quad \text{(True)} \]

**Testing Interval 2:**
For \(y = 0\):
\[ (0 - 1)(0 + 4) = (-1)(4) = -4 < 0 \quad \text{(False)} \]

**Testing Interval 3:**
For \(y = 2\):
\[ (2 - 1)(2 + 4) = (1)(6) = 6 > 0 \quad \text{(True)} \]

### Step 5: Write the solution in terms of \(y\)
The inequality \( (y - 1)(y + 4) > 0 \) holds true for:
\[ y < -4 \quad \text{or} \quad y > 1 \]

### Step 6: Translate back to \(x\)
Since \(y = \tan^{-1}x\), we have:
1. \(\tan^{-1}x < -4\) (not possible since \(\tan^{-1}x\) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\))
2. \(\tan^{-1}x > 1\)

To find the corresponding \(x\) values:
\[ x > \tan(1) \]

### Step 7: Conclusion
Thus, the complete set of values of \(x\) that satisfy the original inequality is:
\[ x > \tan(1) \]