If cos(90° − θ) = x and θ is acute, then what is the value of cos θ in terms of x?

Correct — A. Turn the co-function into sin θ, then use sin² θ + cos² θ = 1.

Complement rule: cos(90° − θ) = sin θ, so sin θ = x
Identity: cos² θ = 1 − sin² θ = 1 − x²
Square root: cos θ = ±√(1 − x²)
θ is acute, so cos θ > 0: cos θ = √(1 − x²) → option (a)

Why the others are wrong

• (b)
  x — x is sin θ, the value the stem gives for cos(90° − θ). Among acute angles, cos θ equals sin θ at θ = 45° alone, so x is not cos θ in general.
• (c)
  1⁄x — 1⁄x is cosec θ, the reciprocal of sin θ. For acute θ, x lies between 0 and 1, so 1⁄x exceeds 1, while cos θ of an acute angle stays below 1.
• (d)
  Undefined — cos θ is defined for every angle, and for acute θ it lies between 0 and 1. The acute condition fixes the sign of the root, nothing more.

Co-function rule: a ratio of θ equals the co-ratio of its complement, 90° − θ.

sin(90° − θ) = cos θ and cos(90° − θ) = sin θ
tan(90° − θ) = cot θ and sec(90° − θ) = cosec θ

Once cos(90° − θ) becomes sin θ, the identity sin² θ + cos² θ = 1 turns it into cos θ, and the acute condition picks the positive root.

Key facts

• cos(90° − θ) = sin θ and sin(90° − θ) = cos θ.
• tan(90° − θ) = cot θ and cot(90° − θ) = tan θ.
• For acute θ, cos θ = √(1 − sin² θ).
• For acute θ, both sin θ and cos θ lie between 0 and 1.

Common traps

• Answering x. That is cos(90° − θ), which the stem gives, not cos θ, which it asks for.
• Keeping ±√(1 − x²). Acute θ makes cos θ positive, so take the positive root.

No directly related past PYQ was found.