If secA + tan A = p, then what is sec A in terms of p?

Correct — A. Use sec²A − tan²A = 1.

Factor it: (sec A − tan A)(sec A + tan A) = 1
Divide by p: sec A − tan A = 1⁄p

Add the two equations: 2 sec A = p + 1⁄p = (p² + 1)⁄p
Halve: sec A = (p² + 1)⁄(2p) → option (a).

Check with A = 0°: p = 1 + 0 = 1, and (1 + 1)⁄2 = 1 = sec 0°.

Why the others are wrong

• (b)
  (p² + 1)⁄(3p) — (p² + 1)⁄(3p) gives 2⁄3 when A = 0° and p = 1, but sec A never lies strictly between −1 and 1. Adding the two equations doubles sec A, so the divisor is 2p.
• (c)
  (p² + 1)⁄(5p) — (p² + 1)⁄(5p) gives 2⁄5 at A = 0°, where sec A = 1. Nothing in the working produces a 5: adding (sec A + tan A) and (sec A − tan A) gives 2 sec A.
• (d)
  (p² + 1)⁄(9p) — (p² + 1)⁄(9p) gives 2⁄9 at A = 0°, where sec A must be 1. The denominator comes from 2 sec A = p + 1⁄p, so it is 2p, not 9p.

sec²A − tan²A = 1 factors as (sec A − tan A)(sec A + tan A) = 1. So whenever sec A + tan A = p, its partner is sec A − tan A = 1⁄p.

Adding the pair isolates sec A. Subtracting isolates tan A: 2 tan A = p − 1⁄p, so tan A = (p² − 1)⁄(2p).

Dividing tan A by sec A gives sin A = (p² − 1)⁄(p² + 1).

Key facts

• (sec A + tan A)(sec A − tan A) = sec²A − tan²A = 1.
• If sec A + tan A = p, then sec A = (p² + 1)⁄(2p) and tan A = (p² − 1)⁄(2p).
• sin A = tan A ÷ sec A = (p² − 1)⁄(p² + 1).
• cosec²A − cot²A = 1 gives the same reciprocal pair for cosec A + cot A.

Common traps

• Squaring sec A + tan A = p, which drags in a 2 sec A tan A term and a much longer route.
• Taking sec A − tan A as −p instead of the reciprocal 1⁄p.

No directly related past PYQ was found.