If sin A = 3⁄5 and cos B = 12⁄13, A and B ∈ (0, π⁄2), find sin(A + B).

Correct — A. Both angles lie in (0, π⁄2), so every ratio is positive.

Complete the 3-4-5 triangle: cos A = √(1 − 9⁄25) = 4⁄5
Complete the 5-12-13 triangle: sin B = √(1 − 144⁄169) = 5⁄13

Expand: sin(A + B) = sin A cos B + cos A sin B
Substitute: (3⁄5)(12⁄13) + (4⁄5)(5⁄13)
Add: 36⁄65 + 20⁄65 = 56⁄65 → option (a)

Why the others are wrong

• (b)
  63⁄65 — 63⁄65 is cos(A − B) = (4⁄5)(12⁄13) + (3⁄5)(5⁄13) = 48⁄65 + 15⁄65. The sine formula pairs a sine with a cosine in each term.
• (c)
  33⁄65 — 33⁄65 is cos(A + B) = cos A cos B − sin A sin B = 48⁄65 − 15⁄65. That is the cosine formula, not the sine one.
• (d)
  77⁄85 — 77⁄85 cannot come from these angles. Every product here has denominator 5 × 13 = 65, and 85 = 5 × 17.

The compound-angle formula is sin(A + B) = sin A cos B + cos A sin B. Each term mixes one sine with one cosine.

To use it you need all four ratios. Build the missing ones from Pythagorean triples: 3-4-5 gives cos A = 4⁄5, and 5-12-13 gives sin B = 5⁄13.

The interval (0, π⁄2) fixes the signs. In the first quadrant sine and cosine are both positive, so both square roots are taken as positive.

Key facts

• sin(A + B) = sin A cos B + cos A sin B.
• cos(A + B) = cos A cos B − sin A sin B, and cos(A − B) = cos A cos B + sin A sin B.
• 3-4-5 and 5-12-13 are Pythagorean triples: 3² + 4² = 5² and 5² + 12² = 13².

Common traps

• Using cos A cos B + sin A sin B, which is cos(A − B) = 63⁄65.
• Subtracting the two terms, which gives sin(A − B) = 36⁄65 − 20⁄65 = 16⁄65.

No directly related past PYQ was found.