If tan θ + cot θ = 2, and θ is an acute angle, then the value of θ is:

Correct — C. Write cot θ as 1⁄tan θ and let t = tan θ.

t + 1⁄t = 2
Multiply through by t: t² + 1 = 2t
t² − 2t + 1 = 0, that is (t − 1)² = 0

So t = 1 — a repeated root, so there is nothing else to check.

tan θ = 1 and θ is acute, which gives θ = 45° → option (c)

Why the others are wrong

• (a)
  15° — At 15°, tan θ = 2 − √3 and cot θ = 2 + √3, so their sum is 4, not 2.
• (b)
  30° — At 30°, tan θ + cot θ = 1⁄√3 + √3 = 4⁄√3 ≈ 2.31, above the required 2.
• (d)
  60° — At 60° the sum is again 4⁄√3 ≈ 2.31 — tan θ + cot θ is symmetric about 45°, so 30° and 60° return identical values.

tan θ + cot θ ≥ 2 for every acute θ, with the minimum of 2 reached exactly where tan θ = cot θ, at 45°. So the equation has a single acute solution.

Algebraically that is the same fact: t + 1⁄t = 2 rearranges to (t − 1)² = 0, a perfect square with a repeated root, which is why the answer is forced rather than chosen.

The expression also simplifies usefully: tan θ + cot θ = 1 ⁄ (sin θ cos θ) = 2 ⁄ sin 2θ. Setting that equal to 2 gives sin 2θ = 1, so 2θ = 90° and θ = 45° again.

Key facts

• t + 1⁄t = 2 has the single root t = 1, because it rearranges to (t − 1)² = 0.
• tan 45° = cot 45° = 1, the one acute angle where the two functions are equal.
• tan θ + cot θ = 1 ⁄ (sin θ cos θ) = 2 ⁄ sin 2θ.
• tan 15° = 2 − √3 and cot 15° = 2 + √3, so their sum is exactly 4.

Common traps

• Treating the quadratic as having two distinct roots when it has a repeated one.
• Reaching tan θ = 1 and then writing 30° or 60° from a half-remembered table.
• Ignoring the acute restriction, since tan θ = 1 also holds at 225°.

No directly related past PYQ was found.