The value of the expression (1 − sin(2t))⁄(1 + sin(2t)) × (cos(t) + sin(t))⁄(cos(t) − sin(t)) is

Correct — B. Both parts of the first fraction are perfect squares once sin 2t is split.

sin 2t = 2 sin t cos t, and sin²t + cos²t = 1, so
1 − sin 2t = cos²t − 2 sin t cos t + sin²t = (cos t − sin t)²
1 + sin 2t = (cos t + sin t)²

The expression becomes
(cos t − sin t)² ⁄ (cos t + sin t)² × (cos t + sin t) ⁄ (cos t − sin t)
= (cos t − sin t) ⁄ (cos t + sin t)

Divide the top and the bottom by cos t:
= (1 − tan t) ⁄ (1 + tan t) → option (b), the second picture.

Why the others are wrong

• (a)
  The 2 belongs to sin 2t, not to tan t. Option (a) keeps a coefficient of 2 on each tangent, but the 2 vanishes the moment 1 ± sin 2t is written as a square, and its signs run the other way too.
• (c)
  Right signs, stray coefficient. Option (c) is the keyed fraction with a 2 stuck to each tangent; dividing (cos t − sin t) by cos t gives plain 1 − tan t, with no number in front.
• (d)
  The two squares swapped. Option (d) is what a sign slip produces — reading 1 − sin 2t as (cos t + sin t)² and 1 + sin 2t as (cos t − sin t)² inverts the whole fraction.

The engine here is the pair of identities 1 ± sin 2t = (cos t ± sin t)². They follow from sin²t + cos²t = 1 together with sin 2t = 2 sin t cos t, and they are what turn an ugly fraction into cancelling factors.

The second move is standard: once an expression is homogeneous in sin t and cos t — every term of the same degree — dividing the numerator and denominator by cos t converts it to tan t.

That is why answers to this family of questions arrive in tan form even though no tangent appears in the question.

Key facts

• sin 2t = 2 sin t cos t.
• 1 − sin 2t = (cos t − sin t)² and 1 + sin 2t = (cos t + sin t)², both by sin²t + cos²t = 1.
• Dividing (cos t − sin t) by cos t gives 1 − tan t.

Common traps

• Expanding everything instead of spotting the perfect square, which leaves a fraction that will not cancel
• Cancelling only one factor of (cos t + sin t) and stopping with a squared denominator
• Never dividing by cos t, so the working stalls before it reaches tan form

No directly related past PYQ was found.