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:
We express everything in terms of sine and cosine and simplify directly without using the identity in the first step.
Step 2: Rewrite p and q:
Given:
p = secα / cscβ
q = tanα / cscβ

Using basic definitions:
secα = 1 / cosα
tanα = sinα / cosα
cscβ = 1 / sinβ

So,
p = (1 / cosα) ÷ (1 / sinβ)
= sinβ / cosα

q = (sinα / cosα) ÷ (1 / sinβ)
= (sinα sinβ) / cosα

Step 3: Compute p² − q²:
p² = sin²β / cos²α
q² = sin²α sin²β / cos²α

So,
p² − q² = [sin²β / cos²α] − [sin²α sin²β / cos²α]
Take common denominator cos²α:
= sin²β (1 − sin²α) / cos²α

But 1 − sin²α = cos²α

Therefore,
p² − q² = sin²β cos²α / cos²α
= sin²β

Step 4: Multiply by sec²α:
LHS = (p² − q²) sec²α
= sin²β × (1 / cos²α)
= sin²β / cos²α

But p = sinβ / cosα
So,
p² = sin²β / cos²α

Hence,
LHS = p² = RHS
Final Answer:
LHS = RHS. Hence proved using an alternative method.