If secA = 13⁄5 and A is acute, find sinA.

Correct — B. Flip sec to get cos, then use sin²A + cos²A = 1.

cos A = 1 ÷ sec A = 5⁄13
sin²A = 1 − (5⁄13)² = 1 − 25⁄169 = 144⁄169
sin A = 12⁄13 (positive, as A is acute) → option (b)

Triangle check: adjacent 5, hypotenuse 13, so opposite = √(169 − 25) = 12 ✓

Why the others are wrong

• (a)
  5⁄13 — 5⁄13 is cos A, the reciprocal of sec A. The question asks for sin A, opposite over hypotenuse, which is 12⁄13.
• (c)
  13⁄12 — 13⁄12 is cosec A, the reciprocal of sin A. It is greater than 1, and sin A never exceeds 1.
• (d)
  13⁄5 — 13⁄5 is sec A itself, copied from the question. Like 13⁄12, it is above 1, which no sine can be.

sec A is the reciprocal of cos A. In a right triangle it is hypotenuse ÷ adjacent, so sec A = 13⁄5 means hypotenuse 13 and adjacent side 5.

Pythagoras gives the third side: √(13² − 5²) = √144 = 12. That is the 5, 12, 13 triple.

Then sin A = opposite ÷ hypotenuse = 12⁄13. The same triangle gives every ratio: tan A = 12⁄5, cot A = 5⁄12, cosec A = 13⁄12.

Key facts

• sec A = 1⁄cos A and cosec A = 1⁄sin A.
• sin²A + cos²A = 1, equivalently sec²A = 1 + tan²A.
• 5, 12, 13 is a Pythagorean triple: 25 + 144 = 169.
• For any angle, −1 ≤ sin A ≤ 1.

Common traps

• Stopping at cos A = 5⁄13, the reciprocal step. The question wants the sine, one Pythagoras step further.
• Flipping the wrong ratio. 13⁄12 is cosec A, the reciprocal of the sine, and any value above 1 cannot be a sine.

No directly related past PYQ was found.