If x = asecθcosØ, y = bsecθsinØ and z = ctanθ, then the value of (x²⁄a² + y²⁄b² − z²⁄c²) is equal to:

Correct — C. Divide each variable by its own constant before squaring, and a, b, c disappear.

x⁄a = secθ cosØ → x²⁄a² = sec²θ cos²Ø
y⁄b = secθ sinØ → y²⁄b² = sec²θ sin²Ø
z⁄c = tanθ → z²⁄c² = tan²θ

Add the first two: sec²θ (cos²Ø + sin²Ø) = sec²θ, because cos²Ø + sin²Ø = 1.

That leaves sec²θ − tan²θ = 1 → option (c).

Why the others are wrong

• (a)
  0 — 0 would need sec²θ = tan²θ. They never coincide: sec²θ − tan²θ equals 1 for every θ at which both are defined.
• (b)
  3 — 3 counts the three terms instead of evaluating them. Two of them merge into sec²θ before anything is subtracted, so three separate values never survive.
• (d)
  2 — 2 adds the two identities together — cos²Ø + sin²Ø = 1 and sec²θ − tan²θ = 1. The first is a factor inside sec²θ, not a separate term to be added on.

Two Pythagorean identities carry the whole question: sin²A + cos²A = 1 and sec²A − tan²A = 1.

The constants a, b and c are camouflage. Dividing x by a, y by b and z by c strips them out, which is why the answer is a bare number rather than an expression in a, b, c.

Note the two different angles, θ and Ø. Only the Ø terms combine with each other, and only the θ terms with each other.

Key facts

• sin²A + cos²A = 1 for every angle A.
• sec²A − tan²A = 1, obtained by dividing sin²A + cos²A = 1 through by cos²A.
• Dividing each variable by its own coefficient before squaring keeps a, b and c out of the algebra entirely.

Common traps

• Squaring first and then hunting for a², b² and c² to cancel by hand.
• Treating θ and Ø as the same angle, which destroys the cos²Ø + sin²Ø pairing.
• Writing sec²θ + tan²θ = 1 — the identity is a difference, not a sum.

No directly related past PYQ was found.