If cos A + cos 2 A = 1 then sin 2 A + sin 4 A is equal to:

Correct — B. The stem prints the powers flat; read it as cos A + cos²A = 1, asking for sin²A + sin⁴A.

Rearrange the given equation:
cos A = 1 − cos²A = sin²A

Substitute that into the target:
sin²A + sin⁴A = sin²A + (sin²A)²
= cos A + cos²A

That is the left side of the given equation again, so the value is 1 → option (b), whose figure reads 1.

Why the others are wrong

• (a)
  Option (a) is 0. That needs sin²A(1 + sin²A) = 0, so sin A = 0 and cos A = ±1 — and then cos A + cos²A is 2 or 0, never the 1 the stem gives.
• (c)
  Option (c) is cos A ⁄ cos²A, which is sec A — still a function of A. The substitution collapses the target onto cos A + cos²A, a fixed number, so no trig ratio can survive.
• (d)
  Option (d) is cos²A ⁄ cos A, which reduces to cos A, and cos A is only sin²A. That is the first term of sin²A + sin⁴A — you stopped one term early.

You are handed a constraint and a target built to fold back onto it. You never solve for A.

The constraint cos A + cos²A = 1 rearranges to cos A = 1 − cos²A, and 1 − cos²A is sin²A by the Pythagorean identity. So cos A = sin²A — one swap that rewrites every sine in the target as a cosine.

The target is sin²A + (sin²A)², which becomes cos A + cos²A, and the constraint has already priced that at 1.

Key facts

• cos A + cos²A = 1 rearranges to cos A = 1 − cos²A, and 1 − cos²A is sin²A.
• sin⁴A is (sin²A)², so the same substitution turns it into cos²A.
• The identity behind the swap is sin²A + cos²A = 1.
• An angle satisfying the constraint does exist — cos A = (√5 − 1) ⁄ 2 ≈ 0.618 — but its value is never needed.

Common traps

• Reading the flattened cos 2 A as cos(2A) rather than cos²A.
• Substituting into the first term only and answering cos A, which is option (d).
• Solving the quadratic in cos A for the angle instead of substituting.

No directly related past PYQ was found.