If `alpha` is a positive acute angle and `2sin alpha+15cos^(2)alpha=7`, then the value of `cotalpha` is

Options

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

Answer

To solve the equation \( 2 \sin \alpha + 15 \cos^2 \alpha = 7 \) for \( \cot \alpha \), we can follow these steps:

### Step 1: Rewrite the equation using the Pythagorean identity
We know that \( \cos^2 \alpha = 1 - \sin^2 \alpha \). We can substitute this into the equation:
\[ 2 \sin \alpha + 15 (1 - \sin^2 \alpha) = 7 \]

### Step 2: Simplify the equation
Expanding the equation gives:
\[ 2 \sin \alpha + 15 - 15 \sin^2 \alpha = 7 \]
Now, rearranging the equation:
\[ -15 \sin^2 \alpha + 2 \sin \alpha + 15 - 7 = 0 \]
This simplifies to:
\[ -15 \sin^2 \alpha + 2 \sin \alpha + 8 = 0 \]
Multiplying through by -1 to make the leading coefficient positive:
\[ 15 \sin^2 \alpha - 2 \sin \alpha - 8 = 0 \]

### Step 3: Solve the quadratic equation
We can use the quadratic formula \( \sin \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 15, b = -2, c = -8 \):
\[ b^2 - 4ac = (-2)^2 - 4 \cdot 15 \cdot (-8) = 4 + 480 = 484 \]
Now substituting into the formula:
\[ \sin \alpha = \frac{-(-2) \pm \sqrt{484}}{2 \cdot 15} = \frac{2 \pm 22}{30} \]
This gives us two potential solutions:
1. \( \sin \alpha = \frac{24}{30} = \frac{4}{5} \)
2. \( \sin \alpha = \frac{-20}{30} = -\frac{2}{3} \) (not valid since \( \alpha \) is acute)

Thus, we have:
\[ \sin \alpha = \frac{4}{5} \]

### Step 4: Find \( \cos \alpha \)
Using the identity \( \cos^2 \alpha = 1 - \sin^2 \alpha \):
\[ \cos^2 \alpha = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \]
Thus,
\[ \cos \alpha = \frac{3}{5} \]

### Step 5: Calculate \( \cot \alpha \)
The cotangent is given by:
\[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Final Answer
The value of \( \cot \alpha \) is:
\[ \cot \alpha = \frac{3}{4} \]

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