If `sectheta-"cosec " theta=(4)/(3)` , then what is `(sin theta-costheta)` equal to ?

Options

A. `-2` only
B. `(1)/(2)` only
C. Both `-2and (1)/(2)`
D. Neither `(1)/(2)"nor "-2`

Explanation

To solve the equation \( \sec \theta - \csc \theta = \frac{4}{3} \) and find the value of \( \sin \theta - \cos \theta \), we can follow these steps:

### Step 1: Convert secant and cosecant to sine and cosine
We know that:
\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \csc \theta = \frac{1}{\sin \theta} \]
Substituting these into the equation gives:
\[ \frac{1}{\cos \theta} - \frac{1}{\sin \theta} = \frac{4}{3} \]

### Step 2: Find a common denominator
The common denominator for the left side is \( \sin \theta \cos \theta \):
\[ \frac{\sin \theta - \cos \theta}{\sin \theta \cos \theta} = \frac{4}{3} \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 3(\sin \theta - \cos \theta) = 4(\sin \theta \cos \theta) \]

### Step 4: Rearrange the equation
Rearranging the equation leads to:
\[ 3\sin \theta - 3\cos \theta = 4\sin \theta \cos \theta \]

### Step 5: Square both sides
To eliminate the terms, we can square both sides:
\[ (3\sin \theta - 3\cos \theta)^2 = (4\sin \theta \cos \theta)^2 \]
Expanding both sides:
\[ 9(\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta) = 16\sin^2 \theta \cos^2 \theta \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 9(1 - 2\sin \theta \cos \theta) = 16\sin^2 \theta \cos^2 \theta \]

### Step 6: Let \( x = \sin \theta \cos \theta \)
Let \( x = \sin \theta \cos \theta \), then:
\[ 9(1 - 2x) = 16x^2 \]
Expanding gives:
\[ 9 - 18x = 16x^2 \]
Rearranging gives:
\[ 16x^2 + 18x - 9 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 16, b = 18, c = -9 \):
\[ x = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 16 \cdot (-9)}}{2 \cdot 16} \]
Calculating the discriminant:
\[ = \frac{-18 \pm \sqrt{324 + 576}}{32} = \frac{-18 \pm \sqrt{900}}{32} = \frac{-18 \pm 30}{32} \]
Calculating the two possible values:
1. \( x = \frac{12}{32} = \frac{3}{8} \)
2. \( x = \frac{-48}{32} = -\frac{3}{2} \) (not valid since \( \sin \theta \) and \( \cos \theta \) cannot be negative)

### Step 8: Substitute back to find \( \sin \theta - \cos \theta \)
Now substituting \( x = \frac{3}{8} \) back into \( \sin \theta \cos \theta \):
\[ \sin \theta \cos \theta = \frac{3}{8} \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) and the relation \( \sin \theta - \cos \theta \):
\[ (\sin \theta - \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta \]
Substituting gives:
\[ (\sin \theta - \cos \theta)^2 = 1 - 2 \cdot \frac{3}{8} = 1 - \frac{3}{4} = \frac{1}{4} \]
Taking the square root:
\[ \sin \theta - \cos \theta = \pm \frac{1}{2} \]

### Final Answer
Thus, the value of \( \sin \theta - \cos \theta \) is:
\[ \sin \theta - \cos \theta = \frac{1}{2} \]