If sinA + cosA = x, then find the value of sin²A + cos²A + 2sinAcosA.
- (a)x
- (b)x²
- (c)1
- (d)x+1
Answer
Why
Correct — B.
Spot the pattern: a² + b² + 2ab = (a + b)²
Here a = sinA and b = cosA
Rewrite: sin²A + cos²A + 2sinAcosA = (sinA + cosA)²
Substitute sinA + cosA = x: (x)² = x²
Cross-check: sin²A + cos²A = 1, so x² = 1 + 2sinAcosA
So the value is x² → option (b)
Why the others are wrong
- (a)x — x is sinA + cosA itself, not this expression. The expression has the form a² + 2ab + b², which is (a + b)², so it equals x².
- (c)1 — 1 is just sin²A + cos²A. It drops the 2sinAcosA term, which is zero only when sinA or cosA is 0.
- (d)x+1 — x + 1 has the wrong extra term. From x² = 1 + 2sinAcosA, the part added to 1 is 2sinAcosA = x² − 1, not x.
Concept
The expression is a perfect square in disguise: a² + 2ab + b² = (a + b)², with a = sinA and b = cosA.
Paired with the Pythagorean identity sin²A + cos²A = 1, the square gives x² = 1 + 2sinAcosA.
From there sinAcosA = (x² − 1)⁄2 and sin2A = 2sinAcosA = x² − 1. That is how a given value of sinA + cosA unlocks the product, the double angle, and higher powers.
Key facts
- (sinA + cosA)² = 1 + 2sinAcosA
- (sinA − cosA)² = 1 − 2sinAcosA
- (sinA + cosA)² + (sinA − cosA)² = 2
- sin2A = 2sinAcosA
Study next
Common traps
- Replacing sin²A + cos²A with 1 and answering 1, which drops the 2sinAcosA term
- Answering x because the stem defines x, without squaring it
22 Sep 2025, 16:00, Quant Q.18 prints the same expression with sinA + cosA = √2⁄2, so the value is (√2⁄2)² = 1⁄2.
17 Sep 2025, 09:00, Quant Q.22 takes the next step: sinA + cosA = 5⁄4 gives sin2A = 25⁄16 − 1 = 9⁄16.
Related PYQs
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