`lim_(x to pi/4)(tan^3x-tanx)/(cos(pi/4+x))=alpha and lim_(x to 0)(cosx)^cotx=beta` .If `alpha and beta` are the roots of equation `ax^2+bx-4` then ordered pair (a,b) is

Answer

To solve the problem, we need to evaluate two limits, \(\alpha\) and \(\beta\), and then find the ordered pair \((a, b)\) such that \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(ax^2 + bx - 4 = 0\).

### Step 1: Calculate \(\alpha = \lim_{x \to \frac{\pi}{4}} \frac{\tan^3 x - \tan x}{\cos\left(\frac{\pi}{4} + x\right)}\)

1. **Substituting \(x = \frac{\pi}{4}\)**:
- We find that \(\tan\left(\frac{\pi}{4}\right) = 1\).
- Thus, the numerator becomes \(\tan^3\left(\frac{\pi}{4}\right) - \tan\left(\frac{\pi}{4}\right) = 1^3 - 1 = 0\).
- The denominator becomes \(\cos\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) = 0\).
- This gives us the indeterminate form \(\frac{0}{0}\).

2. **Applying L'Hôpital's Rule**:
- Differentiate the numerator: \(\frac{d}{dx}(\tan^3 x - \tan x) = 3\tan^2 x \sec^2 x - \sec^2 x\).
- Differentiate the denominator: \(\frac{d}{dx}(\cos(\frac{\pi}{4} + x)) = -\sin(\frac{\pi}{4} + x)\).

3. **Re-evaluate the limit**:
\[ \alpha = \lim_{x \to \frac{\pi}{4}} \frac{3\tan^2 x \sec^2 x - \sec^2 x}{-\sin\left(\frac{\pi}{4} + x\right)} \]
- Substitute \(x = \frac{\pi}{4}\):
\[ = \frac{3(1^2)(\sqrt{2})^2 - (\sqrt{2})^2}{-\sin\left(\frac{\pi}{2}\right)} = \frac{3 \cdot 2 - 2}{-1} = \frac{6 - 2}{-1} = \frac{4}{-1} = -4 \]
- Thus, \(\alpha = -4\).

### Step 2: Calculate \(\beta = \lim_{x \to 0} (\cos x)^{\cot x}\)

1. **Substituting \(x = 0\)**:
- We find that \(\cos(0) = 1\) and \(\cot(0) = \infty\).
- This gives us the indeterminate form \(1^\infty\).

2. **Using the exponential limit**:
\[ \beta = e^{\lim_{x \to 0} (\cos x - 1) \cot x} \]
- We know \(\cot x = \frac{\cos x}{\sin x}\), so:
\[ \beta = e^{\lim_{x \to 0} (\cos x - 1) \frac{\cos x}{\sin x}} \]
- Using the limit \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\):
\[ \beta = e^{\lim_{x \to 0} \frac{-(\cos x - 1)}{x^2} \cdot x} = e^{\lim_{x \to 0} -\frac{1}{2} \cdot 1} = e^{0} = 1 \]
- Thus, \(\beta = 1\).

### Step 3: Find the ordered pair \((a, b)\)

1. **Roots of the quadratic equation**:
- The roots are \(\alpha = -4\) and \(\beta = 1\).
- The sum of the roots \(\alpha + \beta = -4 + 1 = -3\).
- The product of the roots \(\alpha \beta = -4 \cdot 1 = -4\).

2. **Forming the quadratic equation**:
- The quadratic equation can be expressed as:
\[ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \]
- This gives:
\[ x^2 + 3x - 4 = 0 \]

3. **Identifying coefficients**:
- Comparing with \(ax^2 + bx - 4\):
- \(a = 1\)
- \(b = 3\)

### Final Answer
The ordered pair \((a, b)\) is \((1, 3)\).