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 will follow these steps:

### Step 1: Substitute \(\tan^{-1} x\) with a variable
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 look for two numbers that multiply to \(-4\) (the constant term) and add to \(3\) (the coefficient of \(y\)). The numbers \(4\) and \(-1\) work:
\[ y^2 + 4y - 1y - 4 = (y + 4)(y - 1) \]
Thus, we rewrite the inequality as:
\[ (y + 4)(y - 1) > 0 \]

### Step 3: Determine the critical points
The critical points from the factors are:
\[ y + 4 = 0 \quad \Rightarrow \quad y = -4 \]
\[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \]

### Step 4: Test intervals
We will test the intervals determined by the critical points \(-4\) and \(1\):
1. \(y < -4\)
2. \(-4 < y < 1\)
3. \(y > 1\)

- For \(y < -4\) (e.g., \(y = -5\)):
\[ (-5 + 4)(-5 - 1) = (-1)(-6) = 6 > 0 \quad \text{(True)} \]

- For \(-4 < y < 1\) (e.g., \(y = 0\)):
\[ (0 + 4)(0 - 1) = (4)(-1) = -4 < 0 \quad \text{(False)} \]

- For \(y > 1\) (e.g., \(y = 2\)):
\[ (2 + 4)(2 - 1) = (6)(1) = 6 > 0 \quad \text{(True)} \]

### Step 5: Combine results
The inequality \((y + 4)(y - 1) > 0\) holds for:
1. \(y < -4\)
2. \(y > 1\)

### Step 6: Convert back to \(x\)
Now we convert back to \(x\) using \(y = \tan^{-1} x\):
1. For \(y < -4\):
\(\tan^{-1} x < -4\) is not possible since \(\tan^{-1} x\) has a range of \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

2. For \(y > 1\):
\(\tan^{-1} x > 1\) implies:
\[ x > \tan(1) \]
Since \(\tan(1) \approx 1.5574\), we conclude:
\[ x > \tan(1) \]

### Final Solution
The complete set of values of \(x\) that satisfy the inequality is:
\[ x > \tan(1) \]