If `cosectheta-cottheta=q`, then `cosectheta=?`

Options

A. `(1)/(q)`
B. `q+(1)/(q)`
C. `(1)/(2q)`
D. `(1)/(2)(q+(1)/(q))`

Answer

To solve the equation \( \csc \theta - \cot \theta = q \) for \( \csc \theta \), we can follow these steps:

### Step 1: Rewrite the given equation
We start with the equation:
\[ \csc \theta - \cot \theta = q \]

### Step 2: Express \(\csc \theta\) and \(\cot \theta\) in terms of sine and cosine
Recall the definitions:
\[ \csc \theta = \frac{1}{\sin \theta} \quad \text{and} \quad \cot \theta = \frac{\cos \theta}{\sin \theta} \]
Substituting these into the equation gives:
\[ \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} = q \]

### Step 3: Combine the fractions
Since both terms on the left side have a common denominator, we can combine them:
\[ \frac{1 - \cos \theta}{\sin \theta} = q \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 1 - \cos \theta = q \sin \theta \]

### Step 5: Isolate \(\cos \theta\)
Rearranging the equation to isolate \(\cos \theta\):
\[ \cos \theta = 1 - q \sin \theta \]

### Step 6: Use the Pythagorean identity
We know that \(\sin^2 \theta + \cos^2 \theta = 1\). We can substitute \(\cos \theta\) from the previous step into this identity:
\[ \sin^2 \theta + (1 - q \sin \theta)^2 = 1 \]

### Step 7: Expand and simplify
Expanding the squared term:
\[ \sin^2 \theta + (1 - 2q \sin \theta + q^2 \sin^2 \theta) = 1 \]
Combining like terms:
\[ (1 + q^2)\sin^2 \theta - 2q \sin \theta + 1 - 1 = 0 \]
This simplifies to:
\[ (1 + q^2)\sin^2 \theta - 2q \sin \theta = 0 \]

### Step 8: Factor the equation
Factoring out \(\sin \theta\):
\[ \sin \theta \left((1 + q^2)\sin \theta - 2q\right) = 0 \]

### Step 9: Solve for \(\sin \theta\)
Setting each factor to zero gives:
1. \(\sin \theta = 0\) (not valid in this context)
2. \((1 + q^2)\sin \theta - 2q = 0\)

From the second factor:
\[ (1 + q^2)\sin \theta = 2q \]
Thus,
\[ \sin \theta = \frac{2q}{1 + q^2} \]

### Step 10: Substitute back to find \(\csc \theta\)
Now, we can find \(\csc \theta\):
\[ \csc \theta = \frac{1}{\sin \theta} = \frac{1 + q^2}{2q} \]

### Final Answer
Thus, the value of \(\csc \theta\) is:
\[ \csc \theta = \frac{1 + q^2}{2q} \]

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