If `cot alpha` equals the integral solution of inequality `4x ^(2)-16x+15 le 0 and sin beta` equals to the slope of the bisector of the first quadrant, then `sin (alpha+ beta) sin (alpha - beta) ` is equal to:

Options

A. `-3/5`
B. `-4/5`
C. `(2)/(sqrt2)`
D. `3`

Answer

To solve the problem step by step, let's break it down into manageable parts.

### Step 1: Solve the Inequality
We need to solve the inequality \(4x^2 - 16x + 15 \leq 0\).

1. **Find the roots of the equation** \(4x^2 - 16x + 15 = 0\) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
where \(D = b^2 - 4ac\).

Here, \(a = 4\), \(b = -16\), and \(c = 15\).

Calculate \(D\):
\[ D = (-16)^2 - 4 \cdot 4 \cdot 15 = 256 - 240 = 16 \]

Now, substitute into the quadratic formula:
\[ x = \frac{16 \pm \sqrt{16}}{2 \cdot 4} = \frac{16 \pm 4}{8} \]

This gives us two roots:
\[ x_1 = \frac{20}{8} = \frac{5}{2}, \quad x_2 = \frac{12}{8} = \frac{3}{2} \]

### Step 2: Determine the Interval
The roots are \(x = \frac{3}{2}\) and \(x = \frac{5}{2}\). Since the parabola opens upwards (as the coefficient of \(x^2\) is positive), the inequality \(4x^2 - 16x + 15 \leq 0\) holds between the roots:
\[ \frac{3}{2} \leq x \leq \frac{5}{2} \]

### Step 3: Identify Integral Solutions
The integral solutions of the inequality \(4x^2 - 16x + 15 \leq 0\) are the integers in the interval \(\left[\frac{3}{2}, \frac{5}{2}\right]\). The only integer in this interval is:
\[ x = 2 \]
Thus, \( \cot \alpha = 2 \).

### Step 4: Find \(\sin \alpha\)
From \(\cot \alpha = 2\), we can find \(\sin \alpha\):
\[ \cot^2 \alpha = 4 \implies \csc^2 \alpha = 4 + 1 = 5 \]
\[ \sin^2 \alpha = \frac{1}{\csc^2 \alpha} = \frac{1}{5} \]

### Step 5: Find \(\sin \beta\)
The slope of the bisector of the first quadrant is 1 (the line \(y = x\)). Therefore:
\[ \sin \beta = 1 \implies \sin^2 \beta = 1 \]

### Step 6: Calculate \(\sin(\alpha + \beta)\) and \(\sin(\alpha - \beta)\)
Using the identity:
\[ \sin(\alpha + \beta) \sin(\alpha - \beta) = \sin^2 \alpha - \sin^2 \beta \]
Substituting the values we found:
\[ \sin^2 \alpha = \frac{1}{5}, \quad \sin^2 \beta = 1 \]
\[ \sin(\alpha + \beta) \sin(\alpha - \beta) = \frac{1}{5} - 1 = \frac{1}{5} - \frac{5}{5} = -\frac{4}{5} \]

### Final Answer
Thus, the final result is:
\[ \sin(\alpha + \beta) \sin(\alpha - \beta) = -\frac{4}{5} \]