The value of θ, when √3 cos θ + sin θ = 1 (1 ≤ θ ≤ 90°), is:

Correct — D. The stem, printed as an image, is √3 cos θ + sin θ = 1 with 1 ≤ θ ≤ 90°. Divide through by 2, which is √((√3)² + 1²).

(√3⁄2) cos θ + (1⁄2) sin θ = 1⁄2
cos 30° cos θ + sin 30° sin θ = 1⁄2
cos(θ − 30°) = 1⁄2
θ − 30° = +60° or −60°

That gives θ = 90° or θ = −30°, and only 90° lies inside the stated range.

Check: √3 cos 90° + sin 90° = 0 + 1 = 1. So option (d).

Why the others are wrong

• (a)
  60° — At 60° the left side is √3(1⁄2) + (√3⁄2) = √3 ≈ 1.73, not 1.
• (b)
  30° — At 30° it is √3(√3⁄2) + 1⁄2 = 3⁄2 + 1⁄2 = 2, the largest value the expression can take, since 2 cos(θ − 30°) peaks at θ = 30°.
• (c)
  20° — At 20° it is 2 cos(20° − 30°) = 2 cos 10° ≈ 1.97, nowhere near 1.

Any expression a cos θ + b sin θ folds into a single cosine: R cos(θ − α), where R = √(a² + b²) and tan α = b/a.

Here a = √3 and b = 1, so R = 2 and α = 30°, and the whole left side is 2 cos(θ − 30°). The equation becomes cos(θ − 30°) = 1⁄2, a standard angle.

The fold also fixes the range: the expression can take values only between −2 and 2, so setting it to 1 has solutions where setting it to 3 would not.

Key facts

• a cos θ + b sin θ = R cos(θ − α), with R = √(a² + b²).
• For √3 cos θ + sin θ, R = 2, so the expression never leaves the interval −2 to 2.
• cos x = 1⁄2 at x = 60° and at x = −60° between −180° and 180°.
• √3 cos 90° + sin 90° = 1, which is the substitution check for this answer.

Common traps

• Squaring both sides, which brings in a root the range then has to throw away.
• Testing 30° and 60° alone because they are the friendly angles, then picking whichever comes closer to 1.
• Taking θ = −30° from cos(θ − 30°) = 1⁄2 and ignoring the stated range.

No directly related past PYQ was found.