Question

If \(\frac{\sec \alpha}{\csc \beta} = p\) and \(\frac{\tan \alpha}{\csc \beta} = q\), then prove that \((p^2 - q^2) \sec^2 \alpha = p^2\).

Hint:

Isolating common denominators (like \(\csc^2 \beta\)) makes it easier to apply standard trigonometric identities.

Answer 1

Step 1: Understanding the Concept:
Substitute the given values into the expression to be proved and use the identity \(\sec^2 \theta - \tan^2 \theta = 1\).
Step 2: Detailed Explanation:
Given: \(p = \frac{\sec \alpha}{\csc \beta}\) and \(q = \frac{\tan \alpha}{\csc \beta}\).
Calculate \(p^2 - q^2\):
\[ p^2 - q^2 = \frac{\sec^2 \alpha}{\csc^2 \beta} - \frac{\tan^2 \alpha}{\csc^2 \beta} = \frac{\sec^2 \alpha - \tan^2 \alpha}{\csc^2 \beta} \]
Using the identity \(\sec^2 \alpha - \tan^2 \alpha = 1\):
\[ p^2 - q^2 = \frac{1}{\csc^2 \beta} \]
Now, calculate the LHS: \((p^2 - q^2) \sec^2 \alpha\)
\[ LHS = \frac{1}{\csc^2 \beta} \cdot \sec^2 \alpha = \frac{\sec^2 \alpha}{\csc^2 \beta} \]
From the given information, we know that \(\frac{\sec^2 \alpha}{\csc^2 \beta} = p^2\).
Thus, \(LHS = p^2 = RHS\).
Step 4: Final Answer:
LHS = RHS. Hence proved.