If `p="cosec " theta-cottheta and q = ("cosec "theta+cot theta)^(-1)`, then which one of the following is correct ?

Options

A. `pq=1`
B. p = q
C. p+q =1
D. p+q =0

Explanation

To solve the problem, we need to analyze the expressions for \( p \) and \( q \) given in the question.

### Step 1: Write down the expressions for \( p \) and \( q \)
Given:
\[ p = \csc \theta - \cot \theta \]
\[ q = \left( \csc \theta + \cot \theta \right)^{-1} \]

### Step 2: Rewrite \( q \)
We can rewrite \( q \) as:
\[ q = \frac{1}{\csc \theta + \cot \theta} \]

### Step 3: Find the relationship between \( p \) and \( q \)
We can use the identity for the difference of squares:
\[ \csc^2 \theta - \cot^2 \theta = 1 \]
This can be factored as:
\[ (\csc \theta + \cot \theta)(\csc \theta - \cot \theta) = 1 \]
Substituting the expressions for \( p \) and \( q \):
\[ (\csc \theta + \cot \theta)(p) = 1 \]
This implies:
\[ p \cdot (\csc \theta + \cot \theta) = 1 \]

### Step 4: Substitute \( q \) into the equation
From our expression for \( q \), we can substitute:
\[ p \cdot \frac{1}{q} = 1 \]
Thus, we can rearrange this to find:
\[ p = q \]

### Conclusion
The relationship we derived shows that:
\[ p = q \]

### Final Answer
The correct option is:
**B. \( p = q \)**

---