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

Given:
\[ \frac{\sec \alpha}{\csc \beta} = p,\qquad \frac{\tan \alpha}{\csc \beta} = q \]

Step 1: Rewrite the given expressions
\[ p = \sec\alpha \cdot \sin\beta \] \[ q = \tan\alpha \cdot \sin\beta \]

Step 2: Square both equations
\[ p^2 = \sec^2\alpha \cdot \sin^2\beta \] \[ q^2 = \tan^2\alpha \cdot \sin^2\beta \]

Step 3: Form the expression
\[ p^2 - q^2 = \sin^2\beta \left( \sec^2\alpha - \tan^2\alpha \right) \]

Step 4: Use identity
The Pythagorean identity gives:
\[ \sec^2\alpha - \tan^2\alpha = 1 \] Substitute this into the expression:
\[ p^2 - q^2 = \sin^2\beta \]

Step 5: Substitute back for \( p^2 \)
From Step 2: \[ p^2 = \sec^2\alpha \cdot \sin^2\beta \] So, \[ \sin^2\beta = \frac{p^2}{\sec^2\alpha} \]

Step 6: Replace in the equation
\[ p^2 - q^2 = \frac{p^2}{\sec^2\alpha} \] Multiply both sides by \( \sec^2\alpha \): \[ (p^2 - q^2)\sec^2\alpha = p^2 \]

Final Result:
\[ \boxed{(p^2 - q^2)\sec^2\alpha = p^2} \]
Hence proved.