At what angle θ, the value of sin (θ) and cos (θ) will have the same value?

Correct — b. Turn the condition into a single ratio.

sin θ = cos θ
Divide both sides by cos θ, which is non-zero here:
tan θ = 1

The acute angle whose tangent is 1 is 45°.

Check it: sin 45° = 1⁄√2 and cos 45° = 1⁄√2, so the two are equal → option (b).

Why the others are wrong

• (a)
  60° — At 60° the sine is √3⁄2 ≈ 0.866 while the cosine is 1⁄2, so tan 60° = √3, not 1. Past 45° the sine is the larger of the pair.
• (c)
  30° — At 30° the sine is 1⁄2 and the cosine is √3⁄2, so tan 30° = 1⁄√3. Below 45° the cosine leads, so the two cannot be equal here.
• (d)
  75° — 75° separates them further still — tan 75° = 2 + √3 ≈ 3.73. Between 0° and 90° the two graphs cross exactly once, and 75° is past that crossing.

sin θ and cos θ are equal precisely where their ratio is 1, and that ratio is tan θ.

Between 0° and 90° the sine climbs from 0 to 1 while the cosine falls from 1 to 0, so the two graphs cross exactly once — at 45°, where both equal 1⁄√2.

The same fact reads geometrically. In a right triangle the two acute angles are 45° each only when the legs are equal, and it is exactly then that the side opposite an acute angle matches the side adjacent to it.

Key facts

• sin θ = cos θ reduces to tan θ = 1.
• tan 45° = 1, and sin 45° = cos 45° = 1⁄√2.
• In general θ = 45° + 180°n solves it, but only 45° lies between 0° and 90°.

Common traps

• Reading 'the same value' as 'sum to 1' and hunting among 30° and 60°
• Dividing by cos θ without noting that it must be non-zero
• Swapping which of 30° and 60° carries the √3⁄2

No directly related past PYQ was found.