If cot²θ − 2 cos²θ = 0, (0° < θ < 90°), then the value of θ is:

Correct — A. The stem is an image: if cot²θ − 2 cos²θ = 0, with 0° < θ < 90°, find θ.

cot²θ = 2 cos²θ
cos²θ ⁄ sin²θ = 2 cos²θ
cos²θ is non-zero on 0° < θ < 90°, so divide both sides by it:
1 ⁄ sin²θ = 2 → sin²θ = 1⁄2

sin θ = 1⁄√2, so θ = 45° → option (a)

Why the others are wrong

• (b)
  30° — At 30°, cot²30° = 3 and 2cos²30° = 2 × 3⁄4 = 1.5. The left side comes to 1.5, not 0, so 30° fails the equation outright.
• (c)
  90° — 90° does satisfy the algebra — cot 90° = 0 and cos 90° = 0 give 0 − 0 = 0 — but the stem's rider 0° < θ < 90° is an open interval that shuts 90° out.
• (d)
  60° — At 60°, cot²60° = 1⁄3 while 2cos²60° = 2 × 1⁄4 = 1⁄2. The left side is 1⁄3 − 1⁄2 = −1⁄6, negative rather than zero.

A trigonometric equation of this shape reduces the moment everything is written in sine and cosine.

cot θ = cos θ ⁄ sin θ, so cot²θ − 2cos²θ factors as cos²θ (1⁄sin²θ − 2). The factor cos²θ is common to both terms and, on 0° < θ < 90°, is never zero.

So it can be cancelled without losing a root, and what survives is 1⁄sin²θ = 2 — the sine that belongs to 45°.

Key facts

• cot θ = cos θ ⁄ sin θ, so cot²θ = cos²θ ⁄ sin²θ.
• sin 45° = cos 45° = 1⁄√2, so sin²45° = 1⁄2.
• cos 90° = 0, which is exactly why the stem restricts θ before the cancellation.

Common traps

• Cancelling cos²θ without noticing that the range is what makes it legal
• Reading cot²θ as cot of θ² and evaluating the wrong quantity
• Accepting 30° because the residue 1.5 looks close enough to zero

No directly related past PYQ was found.