If `sec x=P, " cosec "x=Q`, then :

Options

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

Answer

To solve the problem where \( \sec x = P \) and \( \csc x = Q \), we can derive a relationship between \( P \) and \( Q \) using trigonometric identities. Here’s a step-by-step solution:

### Step 1: Express \( \sec x \) and \( \csc x \) in terms of \( \cos x \) and \( \sin x \)
We know that:
\[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \csc x = \frac{1}{\sin x} \]
Given \( \sec x = P \), we can write:
\[ \cos x = \frac{1}{P} \]
Similarly, given \( \csc x = Q \), we can write:
\[ \sin x = \frac{1}{Q} \]

### Step 2: Use the Pythagorean identity
We know from trigonometry that:
\[ \sin^2 x + \cos^2 x = 1 \]
Substituting the expressions for \( \sin x \) and \( \cos x \):
\[ \left(\frac{1}{Q}\right)^2 + \left(\frac{1}{P}\right)^2 = 1 \]

### Step 3: Simplify the equation
This gives us:
\[ \frac{1}{Q^2} + \frac{1}{P^2} = 1 \]
To combine these fractions, we need a common denominator:
\[ \frac{P^2 + Q^2}{P^2 Q^2} = 1 \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ P^2 + Q^2 = P^2 Q^2 \]

### Conclusion
Thus, we have derived the relationship:
\[ P^2 + Q^2 = P^2 Q^2 \]
This is the required relation.