If sinA = 5⁄13 and A lies in 2nd quadrant, then find the value of sinA × secA.

Correct — B. sec A = 1 ⁄ cos A, so sin A × sec A = sin A ⁄ cos A = tan A.

Find cos²A: 1 − sin²A = 1 − 25⁄169 = 144⁄169
Cosine is negative in the 2nd quadrant: cos A = −12⁄13
So sec A = −13⁄12

Multiply: 5⁄13 × (−13⁄12) = −5⁄12 → option (b)

Why the others are wrong

• (a)
  5⁄12 — Right size, wrong sign. In the 2nd quadrant cos A is negative, so sec A = −13⁄12 and the product is −5⁄12. A value of +5⁄12 would need A in the 1st quadrant.
• (c)
  25⁄12 — The 13s cancel, leaving 5 on top, not 25: 5⁄13 × (−13⁄12) = −5⁄12. 25⁄12 is also positive, while tan A is negative in the 2nd quadrant.
• (d)
  25⁄13 — 25⁄13 is positive, but tan A is negative in the 2nd quadrant. Its size is off too: sin A ⁄ cos A has magnitude (5⁄13) ⁄ (12⁄13) = 5⁄12.

sin A × sec A is tan A in disguise, because sec A = 1 ⁄ cos A. The question tests two things: the 5-12-13 triangle and the sign of cosine.

Signs by quadrant follow ASTC: all six ratios positive in the 1st, only sin and cosec in the 2nd, only tan and cot in the 3rd, only cos and sec in the 4th.

In the 2nd quadrant sin A > 0 and cos A < 0, so tan A < 0.

Key facts

• sin²A + cos²A = 1.
• 5² + 12² = 25 + 144 = 169 = 13², so 5-12-13 is a Pythagorean triple.
• In the 2nd quadrant (90° < A < 180°), sin and cosec are positive, while cos, sec, tan and cot are negative.
• sec A = 1 ⁄ cos A, so sin A × sec A = tan A.

Common traps

• Taking cos A = +12⁄13 because triangle sides are positive: the triangle gives the size, the quadrant gives the sign.
• Writing sec A as 1 ⁄ sin A: that is cosec A, and sin A × cosec A = 1.

No directly related past PYQ was found.