If `8sectheta+6cosectheta=20`, then `cottheta=?`

Options

A. `(3)/(4)`
B. `(4)/(3)`
C. none of these
D. cant determine

Answer

To solve the equation \( 8 \sec \theta + 6 \csc \theta = 20 \) and find \( \cot \theta \), we can follow these steps:

### Step 1: Rewrite the secant and cosecant in terms of sine and cosine
We know that:
\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \csc \theta = \frac{1}{\sin \theta} \]
Thus, we can rewrite the equation as:
\[ 8 \cdot \frac{1}{\cos \theta} + 6 \cdot \frac{1}{\sin \theta} = 20 \]

### Step 2: Multiply through by \(\sin \theta \cos \theta\)
To eliminate the fractions, we multiply the entire equation by \(\sin \theta \cos \theta\):
\[ 8 \sin \theta + 6 \cos \theta = 20 \sin \theta \cos \theta \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ 20 \sin \theta \cos \theta - 8 \sin \theta - 6 \cos \theta = 0 \]

### Step 4: Factor the equation
This equation can be factored or solved using substitution. We can express \(\sin \theta\) in terms of \(\cos \theta\) or vice versa. For simplicity, let's express \(\sin \theta\) in terms of \(\cos \theta\):
Let \( \sin \theta = x \) and \( \cos \theta = \sqrt{1 - x^2} \). Substituting this into the equation can be complex, so we will use a trigonometric identity instead.

### Step 5: Use a right triangle
Using a Pythagorean triplet, we can assume:
- Opposite side (perpendicular) = 3
- Adjacent side (base) = 4
- Hypotenuse = 5 (from \(3^2 + 4^2 = 5^2\))

### Step 6: Calculate secant and cosecant
From the triangle:
\[ \sec \theta = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{5}{4} \]
\[ \csc \theta = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{5}{3} \]

### Step 7: Substitute back into the equation
Substituting these values into the original equation:
\[ 8 \cdot \frac{5}{4} + 6 \cdot \frac{5}{3} = 20 \]

Calculating each term:
\[ 10 + 10 = 20 \]
This confirms our values are correct.

### Step 8: Find cotangent
Now, we can find \( \cot \theta \):
\[ \cot \theta = \frac{\text{Base}}{\text{Perpendicular}} = \frac{4}{3} \]

Thus, the final answer is:
\[ \cot \theta = \frac{4}{3} \]

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