If x = p secA cos B, y = q secA sinB and z = r tanA, what is the value of the following expression? x²⁄p² + y²⁄q² − z²⁄r²

Correct — D. The stem is an image: x = p secA cos B, y = q secA sinB, z = r tanA, and it asks for x²⁄p² + y²⁄q² − z²⁄r². Option (d) shows 1.

x⁄p = secA cos B, y⁄q = secA sinB, z⁄r = tanA
x²⁄p² + y²⁄q² = sec²A (cos²B + sin²B)
cos²B + sin²B = 1, so those two terms come to sec²A

sec²A − tan²A = 1 → option (d)

Why the others are wrong

• (a)
  Option (a) keeps p, q and r in the answer. Each one divides out the moment you form x⁄p, y⁄q and z⁄r, so no expression carrying them can survive.
• (b)
  Option (b) is 0, which would need sec²A = tan²A. Those two differ by exactly 1 for every A, since sec²A = 1 + tan²A, so the expression never vanishes.
• (c)
  Option (c) also keeps the parameters, and it reverses the printed signs. The stem adds y²⁄q² and subtracts z²⁄r², which is what makes sec²A − tan²A appear.

Two Pythagorean identities run this question, one after the other.

cos²B + sin²B = 1 clears the angle B, leaving sec²A.
sec²A − tan²A = 1 then clears A as well.

The letters p, q and r are scaffolding. Dividing x by p, y by q and z by r strips them off before any trigonometry starts. That is the design: a stem crowded with six letters that collapses to a bare number.

Key facts

• cos²B + sin²B = 1 for every angle B.
• sec²A − tan²A = 1, which follows from dividing sin²A + cos²A = 1 through by cos²A.
• x⁄p = secA cos B and y⁄q = secA sinB, so x²⁄p² + y²⁄q² = sec²A.

Common traps

• Choosing an option that still contains p, q or r
• Writing sec²A − tan²A = 0 by false analogy with sin²A + cos²A = 1
• Expanding secA into 1⁄cosA at the start, which hides the cos²B + sin²B grouping

No directly related past PYQ was found.