Question

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

Hint:

Isolating the common denominator in substitution problems often makes the algebra much cleaner.

Answer 1

Step 1: Understanding the Concept:
The goal is to substitute the given values of \( p \) and \( q \) into the expression and use the identity \( \sec^2 \alpha - \tan^2 \alpha = 1 \).
Step 2: Key Formula or Approach:
Trigonometric Identity: \( \sec^2 \alpha - \tan^2 \alpha = 1 \).
Step 3: Detailed Explanation:
Start with the term \( (p^2 - q^2) \):
\[ p^2 - q^2 = \left( \frac{\sec \alpha}{\text{cosec} \beta} \right)^2 - \left( \frac{\tan \alpha}{\text{cosec} \beta} \right)^2 \]
\[ p^2 - q^2 = \frac{\sec^2 \alpha - \tan^2 \alpha}{\text{cosec}^2 \beta} \]
Using the identity \( \sec^2 \alpha - \tan^2 \alpha = 1 \):
\[ p^2 - q^2 = \frac{1}{\text{cosec}^2 \beta} \]
Now, substitute this into the LHS of the equation to be proved:
\[ \text{LHS} = (p^2 - q^2) \sec^2 \alpha \]
\[ \text{LHS} = \frac{1}{\text{cosec}^2 \beta} \cdot \sec^2 \alpha \]
\[ \text{LHS} = \frac{\sec^2 \alpha}{\text{cosec}^2 \beta} \]
Since \( p = \frac{\sec \alpha}{\text{cosec} \beta} \), then \( p^2 = \frac{\sec^2 \alpha}{\text{cosec}^2 \beta} \).
Thus, LHS \( = p^2 \), which is the RHS.
Step 4: Final Answer:
LHS \( = \) RHS. Hence Proved.