If sin A + cos A = 1, then find the value of sin⁴A + cos⁴A.

Correct — A.
Square the given sum: (sin A + cos A)² = 1² = 1
Expand: sin²A + cos²A + 2 sin A cos A = 1
Use sin²A + cos²A = 1: 1 + 2 sin A cos A = 1, so sin A cos A = 0

Rewrite the target: sin⁴A + cos⁴A = (sin²A + cos²A)² − 2 sin²A cos²A
Substitute: = 1² − 2 × 0²
= 1 → option (a)

Why the others are wrong

• (b)
  1⁄2 — 1⁄2 is the value at A = 45°, where each fourth power is (1⁄√2)⁴ = 1⁄4. But at 45°, sin A + cos A = √2, not 1, so the given condition rules that angle out.
• (c)
  3⁄4 — 3⁄4 would need 1 − 2 sin²A cos²A = 3⁄4, that is sin²A cos²A = 1⁄8. Squaring the given sum forces sin A cos A = 0, so this value cannot occur here.
• (d)
  5⁄8 — 5⁄8 is the value at A = 30° or 60°: 1⁄16 + 9⁄16 = 10⁄16. There sin A + cos A = (1 + √3)⁄2 ≈ 1.37, not 1.

A sum of sin and cos opens up when you square it. Because sin²A + cos²A = 1, the square always leaves 1 + 2 sin A cos A, so a given value of sin A + cos A fixes the product sin A cos A.

Fourth powers come from the same identity squared: sin⁴A + cos⁴A = 1 − 2 sin²A cos²A. The chain runs sum → product → fourth powers, and you never need to find A itself.

Key facts

• (sin A + cos A)² = 1 + 2 sin A cos A
• sin⁴A + cos⁴A = 1 − 2 sin²A cos²A
• sin⁴A + cos⁴A always lies between 1⁄2 (at A = 45°) and 1 (when sin A cos A = 0)

Common traps

• Plugging in A = 45° or A = 30° out of habit, which gives 1⁄2 or 5⁄8 for angles that do not satisfy sin A + cos A = 1.
• Dropping the − 2 sin²A cos²A term. It vanishes here because sin A cos A = 0, but it decides the answer when the given sum is, say, 1⁄(2√2).

No directly related past PYQ was found.