Ifsin A = m⁄n, then what is the value of (1 + tan²A)?

Correct — A.
Identity: 1 + tan²A = sec²A = 1⁄cos²A
cos²A = 1 − sin²A = 1 − m²⁄n² = (n² − m²)⁄n²
Invert: 1 + tan²A = n²⁄(n² − m²) → option (a)

Check with m = 3, n = 5: tan A = 3⁄4, so 1 + 9⁄16 = 25⁄16
and n²⁄(n² − m²) = 25⁄(25 − 9) = 25⁄16

Why the others are wrong

• (b)
  2n²⁄(n²−m²) — 2n²⁄(n² − m²) is twice the value. Put m = 0: sin A = 0, so tan A = 0 and 1 + tan²A = 1, but this option gives 2.
• (c)
  n²⁄(2(n²−m²)) — n²⁄(2(n² − m²)) is half the value. At m = 0 it gives 1⁄2, while 1 + tan²A = 1 there.
• (d)
  5n²⁄(n²−m²) — 5n²⁄(n² − m²) is five times the value. At m = 3, n = 5 it gives 125⁄16, while 1 + tan²A is 25⁄16.

Divide sin²A + cos²A = 1 by cos²A and you get 1 + tan²A = sec²A. So the expression is just 1⁄cos²A.

Given sin A = m⁄n, take the side opposite A as m and the hypotenuse as n. The adjacent side is then √(n² − m²), so cos²A = (n² − m²)⁄n², and inverting it gives the answer.

Key facts

• 1 + tan²A = sec²A.
• 1 + cot²A = cosec²A.
• If sin A = m⁄n, then cos²A = (n² − m²)⁄n².

Common traps

• Using 1 + tan²A = cosec²A instead, which gives n²⁄m², not offered.
• Writing cos²A as n² − m² and forgetting to divide by n².

No directly related past PYQ was found.