Directions (FOR NEXT TWO): Let `alpha` be the root of the equation `25cos^2theta + 5costheta - 12 = 0`, where `(pi)/(2) lt alpha lt pi`.
What is the `tan alpha` equal to?

Options

A. `-3/4`
B. `3/4`
C. `-4/3`
D. `-4/5`

Answer

To solve the problem, we need to find the value of \( \tan \alpha \) where \( \alpha \) is a root of the equation:

\[ 25 \cos^2 \theta + 5 \cos \theta - 12 = 0 \]

with the condition that \( \frac{\pi}{2} < \alpha < \pi \).

### Step 1: Substitute \( \theta \) with \( \alpha \)

We start by substituting \( \theta \) with \( \alpha \):

\[ 25 \cos^2 \alpha + 5 \cos \alpha - 12 = 0 \]

### Step 2: Factor the quadratic equation

Next, we will factor the quadratic equation. We need two numbers that multiply to \( 25 \times (-12) = -300 \) and add to \( 5 \). The numbers \( 20 \) and \( -15 \) work:

\[ 25 \cos^2 \alpha + 20 \cos \alpha - 15 \cos \alpha - 12 = 0 \]

Rearranging gives:

\[ (25 \cos^2 \alpha + 20 \cos \alpha) + (-15 \cos \alpha - 12) = 0 \]

Factoring by grouping:

\[ 5 \cos \alpha (5 \cos \alpha + 4) - 3(5 \cos \alpha + 4) = 0 \]

This can be factored as:

\[ (5 \cos \alpha - 3)(5 \cos \alpha + 4) = 0 \]

### Step 3: Solve for \( \cos \alpha \)

Setting each factor to zero gives us:

1. \( 5 \cos \alpha - 3 = 0 \) → \( \cos \alpha = \frac{3}{5} \)
2. \( 5 \cos \alpha + 4 = 0 \) → \( \cos \alpha = -\frac{4}{5} \)

Since \( \alpha \) is in the second quadrant (where cosine is negative), we take:

\[ \cos \alpha = -\frac{4}{5} \]

### Step 4: Find \( \sin \alpha \)

Now, we can find \( \sin \alpha \) using the Pythagorean identity:

\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]

Substituting \( \cos \alpha \):

\[ \sin^2 \alpha + \left(-\frac{4}{5}\right)^2 = 1 \]

Calculating \( \left(-\frac{4}{5}\right)^2 \):

\[ \sin^2 \alpha + \frac{16}{25} = 1 \]

Subtracting \( \frac{16}{25} \) from both sides:

\[ \sin^2 \alpha = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \]

Taking the square root:

\[ \sin \alpha = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

Since \( \alpha \) is in the second quadrant, \( \sin \alpha \) is positive:

\[ \sin \alpha = \frac{3}{5} \]

### Step 5: Calculate \( \tan \alpha \)

Now we can find \( \tan \alpha \):

\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{5} \cdot -\frac{5}{4} = -\frac{3}{4} \]

Thus, the value of \( \tan \alpha \) is:

\[ \tan \alpha = -\frac{3}{4} \]

### Final Answer

\[ \text{The value of } \tan \alpha \text{ is } -\frac{3}{4}. \]