If (sin θ)⁄x = (cos θ)⁄y, then sin θ − cos θ is:

Correct — A. Set both ratios equal to one constant k.

sin θ⁄x = cos θ⁄y = k
So sin θ = kx and cos θ = ky

Use sin²θ + cos²θ = 1: k²x² + k²y² = 1
So k² = 1⁄(x² + y²)
Take the positive root: k = 1⁄√(x² + y²)

sin θ − cos θ = kx − ky = k(x − y)
= (x − y)⁄√(x² + y²) → option (a)

Why the others are wrong

• (b)
  (5x−y)⁄(3√(x²+y²)) — (5x − y)⁄(3√(x² + y²)) is not k(x − y). Test x = 3, y = 4, where sin θ = 3⁄5, cos θ = 4⁄5 and the answer is −1⁄5: it gives 11⁄15.
• (c)
  (x−y)⁄(2√(x²+y²)) — The extra 2 halves the answer. At x = 3, y = 4, sin θ − cos θ = 3⁄5 − 4⁄5 = −1⁄5, but this option gives −1⁄10.
• (d)
  (x−y)⁄√(3x²+5y²) — √(3x² + 5y²) is not √(x² + y²), which sin²θ + cos²θ = 1 forces. At x = 3, y = 4 it gives −1⁄√107, not −1⁄5.

When two ratios are equal, set them equal to one constant k. That writes sin θ and cos θ as kx and ky, and the identity sin²θ + cos²θ = 1 then fixes k.

The triangle view gives the same result. Dividing the two ratios gives tan θ = x⁄y, so draw a right triangle with side x opposite θ, side y adjacent and hypotenuse √(x² + y²).

Then sin θ = x⁄√(x² + y²) and cos θ = y⁄√(x² + y²).

Key facts

• If a⁄b = c⁄d = k, then a = kb and c = kd.
• sin²θ + cos²θ = 1 for every angle θ.
• sin θ⁄x = cos θ⁄y is the same as tan θ = x⁄y.

Common traps

• Stopping at k² = 1⁄(x² + y²) and forgetting the square root.
• Writing tan θ = y⁄x: sin θ sits over x, so x is the side opposite θ.

No directly related past PYQ was found.