Simplify (1 + sin θ − cos θ) ⁄ (1 + sin θ + cos θ).

Correct — A. Pair the 1 with the cosine, not with the sine, and the half-angle identities do the rest. Write t = θ⁄2.

1 − cos θ = 2 sin²t · 1 + cos θ = 2 cos²t · sin θ = 2 sin t cos t

Numerator = (1 − cos θ) + sin θ = 2 sin²t + 2 sin t cos t = 2 sin t (sin t + cos t)
Denominator = (1 + cos θ) + sin θ = 2 cos²t + 2 sin t cos t = 2 cos t (sin t + cos t)

The 2 and the bracket (sin t + cos t) are common to both, so they cancel.
What is left is sin t ⁄ cos t = tan(θ⁄2) → option (a).

Why the others are wrong

• (b)
  Option (b) is cos(θ⁄2), which never exceeds 1. As θ approaches 180°, cos θ → −1 and the given fraction runs to 2 ⁄ 0, growing without bound, so no cosine can equal it.
• (c)
  Option (c) is cot(θ⁄2), the reciprocal — what you get by cancelling and then reading cos t over sin t. At θ = 60° the expression is 1.366 ⁄ 2.366 = 0.577, while cot 30° = 1.732.
• (d)
  Option (d) is sin(θ⁄2), which is also capped at 1 and is too small in the middle of the range: at θ = 60° the expression is 0.577 but sin 30° = 0.5.

Every term here has a half-angle form. The three identities that matter are 1 − cos θ = 2 sin²(θ⁄2), 1 + cos θ = 2 cos²(θ⁄2) and sin θ = 2 sin(θ⁄2) cos(θ⁄2).

The move that makes them usable is the regrouping: read the numerator as (1 − cos θ) + sin θ and the denominator as (1 + cos θ) + sin θ. Pair the 1 with the sine instead and nothing factors.

After the substitution both lines carry the factor (sin t + cos t), which cancels. That is why the answer is a clean single function rather than an expression — a good sign you have grouped correctly.

Key facts

• 1 − cos θ = 2 sin²(θ⁄2), the identity that turns the numerator into a sine factor.
• 1 + cos θ = 2 cos²(θ⁄2), the matching identity for the denominator.
• sin θ = 2 sin(θ⁄2) cos(θ⁄2), which supplies the shared bracket that cancels.
• At θ = 60° the whole expression equals 0.577, which is tan 30°.

Common traps

• Grouping the 1 with sin θ, after which neither line factors.
• Cancelling correctly but inverting the last ratio, which lands on cot(θ⁄2).
• Checking the answer only at θ = 90°, where tangent and cotangent of the half-angle are both 1.

No directly related past PYQ was found.