If secx=P, cosecx=Q, then:

Options

A. `P^2+Q^2=PQ`
B. `P^2+Q^2=P^2Q^2`
C. `P^2-Q^2=P^2Q^2`
D. `P^2+Q^2=-P^2Q^2`

Answer

To solve the problem where sec(x) = P and cosec(x) = Q, we can derive a relationship between P and Q step by step.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- We know that:
\[ \sec(x) = \frac{1}{\cos(x)} \quad \text{and} \quad \csc(x) = \frac{1}{\sin(x)} \]
- Therefore, we can express P and Q as:
\[ P = \frac{1}{\cos(x)} \quad \text{and} \quad Q = \frac{1}{\sin(x)} \]

2. **Squaring Both Sides**:
- We square both P and Q:
\[ P^2 = \frac{1}{\cos^2(x)} \quad \text{and} \quad Q^2 = \frac{1}{\sin^2(x)} \]

3. **Adding the Two Equations**:
- Now, we add P² and Q²:
\[ P^2 + Q^2 = \frac{1}{\cos^2(x)} + \frac{1}{\sin^2(x)} \]

4. **Finding a Common Denominator**:
- The common denominator for the right side is \(\sin^2(x) \cos^2(x)\):
\[ P^2 + Q^2 = \frac{\sin^2(x) + \cos^2(x)}{\sin^2(x) \cos^2(x)} \]

5. **Using the Pythagorean Identity**:
- We know from trigonometric identities that:
\[ \sin^2(x) + \cos^2(x) = 1 \]
- Substituting this into our equation gives:
\[ P^2 + Q^2 = \frac{1}{\sin^2(x) \cos^2(x)} \]

6. **Rearranging the Equation**:
- We can rewrite this as:
\[ P^2 + Q^2 = \sec^2(x) \csc^2(x) \]

7. **Conclusion**:
- Thus, we have derived that:
\[ P^2 + Q^2 = \sec^2(x) \csc^2(x) \]

### Final Result:
The relationship between P and Q is:
\[ P^2 + Q^2 = \sec^2(x) \csc^2(x) \]