If sin θ + cos θ = √3 cos θ, then the value of cot θ is:

Correct — C. The stem is a picture: if sin θ + cos θ = √3 cos θ, find cot θ.

Move the cos θ terms to one side:
sin θ = √3 cos θ − cos θ = (√3 − 1) cos θ

So cos θ ⁄ sin θ = 1 ⁄ (√3 − 1), and cos θ ⁄ sin θ is cot θ.

Rationalise with the conjugate (√3 + 1):
cot θ = (√3 + 1) ⁄ ((√3)² − 1²) = (√3 + 1) ⁄ 2

cot θ = (√3 + 1)⁄2, the value pictured in option (c).

Why the others are wrong

• (a)
  Option (a) shows √3 + 1, the rationalised numerator with the denominator thrown away. (√3 − 1)(√3 + 1) = 3 − 1 = 2, and that 2 has to stay under the line.
• (b)
  Option (b) shows √3 − 1, which is tan θ. The relation gives sin θ ⁄ cos θ = √3 − 1, and the question asks for its reciprocal.
• (d)
  Option (d) shows (√3 − 1)⁄2. The denominator is right but the sign inside is not: rationalising 1⁄(√3 − 1) multiplies by the conjugate (√3 + 1), so the numerator gains a plus.

A single linear relation between sin θ and cos θ is enough to fix tan θ, and with it cot θ, because dividing through by cos θ leaves a pure ratio and the angle is never needed.

Here collecting cos θ gives sin θ = (√3 − 1) cos θ, so tan θ = √3 − 1 and cot θ is its reciprocal.

The second half is surd handling: 1⁄(√3 − 1) is cleared by multiplying above and below by the conjugate (√3 + 1), and (√3)² − 1² = 2.

Key facts

• sin θ + cos θ = √3 cos θ rearranges to sin θ = (√3 − 1) cos θ, so tan θ = √3 − 1.
• cot θ is the reciprocal of tan θ, so cot θ = 1⁄(√3 − 1).
• Rationalising with the conjugate gives (√3 + 1)⁄2, because (√3)² − 1² = 2.
• Standard surd values worth holding: cot 15° = 2 + √3 and cot 75° = 2 − √3, reached by the same rationalising step.

Common traps

• Reporting tan θ when the question asks cot θ, which is exactly the value option (b) pictures.
• Squaring both sides to bring in sin²θ + cos²θ = 1. It works, but squaring loses the sign and admits −(√3 − 1) as a second root.
• Rationalising with (√3 − 1) instead of the conjugate, which leaves the surd in place rather than clearing it.

No directly related past PYQ was found.