Find the value of [(4 cos(90 − θ) sin³(90 + θ) − 4 sin(90 + θ) cos³(90 − θ)) ⁄ cos((180 + 8θ) ⁄ 2)].

Correct — A.

The stem is an image. It asks for the value of [4 cos(90 − θ) sin³(90 + θ) − 4 sin(90 + θ) cos³(90 − θ)] ⁄ cos((180 + 8θ) ⁄ 2).

Convert the allied angles first:
cos(90 − θ) = sin θ
sin(90 + θ) = cos θ

Numerator = 4 sin θ cos³θ − 4 cos θ sin³θ
= 4 sin θ cos θ (cos²θ − sin²θ)
= 2 × (2 sin θ cos θ) × cos 2θ
= 2 sin 2θ cos 2θ = sin 4θ

Denominator = cos((180 + 8θ) ⁄ 2) = cos(90 + 4θ) = −sin 4θ

sin 4θ ⁄ (−sin 4θ) = −1 → option (a).

Why the others are wrong

• (b)
  1 — 1 is what you get by reducing cos(90 + 4θ) to +sin 4θ. The correct reduction is cos(90° + x) = −sin x, and that one minus sign is the entire question.
• (c)
  2 — 2 needs two slips at once. Writing 2 sin 2θ cos 2θ as 2 sin 4θ keeps a factor already spent — it equals sin 4θ — and that slip alone gives −2. Only losing the minus in cos(90° + 4θ) as well turns it into +2.
• (d)
  0 — 0 would need the numerator to vanish. It does vanish where sin 4θ = 0 — but the denominator is −sin 4θ, so it vanishes at exactly the same angles, leaving the expression undefined there rather than 0.

Two mechanisms, applied in order.

First, allied angles: adding or subtracting 90° swaps sine for cosine, and the sign follows the quadrant. So cos(90 − θ) = sin θ, sin(90 + θ) = cos θ, and cos(90 + θ) = −sin θ.

Second, double angles, used twice. Pulling out 4 sin θ cos θ leaves cos²θ − sin²θ, which is cos 2θ, so the numerator is 2 sin 2θ cos 2θ — and that is sin 4θ.

The denominator halves to 90 + 4θ, whose cosine is −sin 4θ. The sin 4θ cancels and the value is a constant, which is why no θ was ever supplied.

Key facts

• cos(90° − θ) = sin θ and sin(90° + θ) = cos θ.
• cos(90° + x) = −sin x, which is the source of the minus sign in the answer.
• sin 2x = 2 sin x cos x and cos 2x = cos²x − sin²x, applied in turn, fold 4 sin θ cos θ (cos²θ − sin²θ) into sin 4θ.
• The expression equals −1 at every θ where it is defined, that is wherever sin 4θ ≠ 0.

Common traps

• Taking sin(90° + θ) as −cos θ, when the sine of a second-quadrant angle is positive.
• Reading cos((180 + 8θ) ⁄ 2) as cos(180 + 4θ) by halving only the 8θ.
• Cancelling sin 4θ against −sin 4θ and reporting 1 instead of −1.

No directly related past PYQ was found.