If sinA = 3⁄5 and A lies in the second quadrant, find the value of (sinA + cosA)².

Correct — C.
cos²A = 1 − sin²A = 1 − 9⁄25 = 16⁄25, so cos A = ±4⁄5
Second quadrant: cosine is negative, so cos A = −4⁄5
Add: sin A + cos A = 3⁄5 − 4⁄5 = −1⁄5

Square it: (−1⁄5)² = 1⁄25 → option (c)
Check: 1 + 2 sin A cos A = 1 + 2(3⁄5)(−4⁄5) = 1 − 24⁄25 = 1⁄25

Why the others are wrong

• (a)
  0 — 0 needs sin A = −cos A, which in the second quadrant happens only at A = 135°, where both have size 1⁄√2. Here the sizes are 3⁄5 and 4⁄5, so the sum is −1⁄5.
• (b)
  1 — 1 is sin²A + cos²A. Squaring a sum also adds the cross term 2 sin A cos A = −24⁄25, which brings the value down to 1⁄25.
• (d)
  1⁄5 — 1⁄5 is the size of sin A + cos A = −1⁄5 before squaring. The question asks for the square: (−1⁄5)² = 1⁄25.

Given one ratio, the Pythagorean identity sin²A + cos²A = 1 gives the other ratio's size, but not its sign.

The quadrant supplies the sign. In the second quadrant (90° < A < 180°) sine is positive and cosine is negative, so cos A = −4⁄5, not +4⁄5.

Key facts

• Second quadrant (90° to 180°): sine is positive, cosine and tangent are negative.
• (sin A + cos A)² = 1 + 2 sin A cos A.
• sin A = 3⁄5 pairs with cos A = ±4⁄5, from the 3-4-5 right triangle.

Common traps

• Taking cos A = +4⁄5 by default, which gives 49⁄25.
• Squaring term by term to get sin²A + cos²A = 1 and dropping the cross term.

No directly related past PYQ was found.