Ifcot A = x + 1⁄x, find cosec²A.

Correct — C. Use cosec²A = 1 + cot²A, and expand the square in full.

cot²A = (x + 1⁄x)² = x² + 2·x·(1⁄x) + 1⁄x²
= x² + 2 + 1⁄x²
cosec²A = 1 + x² + 2 + 1⁄x²
= x² + 1⁄x² + 3 → option (c)

Why the others are wrong

• (a)
  x² + 1⁄x² + 1 — x² + 1⁄x² + 1 drops the cross term 2 from (x + 1⁄x)². It is also what cot²A − 1 gives, which flips the sign in the identity.
• (b)
  x² + 1⁄x² + 1 — Printed exactly like option (a) on the response sheet, so it fails for the same reason: the expansion of (x + 1⁄x)² has lost its cross term 2.
• (d)
  x² + 1⁄x² − 1 — x² + 1⁄x² − 1 flips the identity to cosec²A = cot²A − 1 and drops the cross term 2. Each slip on its own gives x² + 1⁄x² + 1 instead.

Divide sin²A + cos²A = 1 through by sin²A and you get 1 + cot²A = cosec²A. Divide by cos²A instead and you get 1 + tan²A = sec²A.

The algebra step is the square of a sum: (x + 1⁄x)² = x² + 2 + 1⁄x², because the cross term 2·x·(1⁄x) is exactly 2.

The identity adds 1 to that 2, which is where the +3 comes from.

Key facts

• 1 + cot²A = cosec²A.
• 1 + tan²A = sec²A.
• (x + 1⁄x)² = x² + 1⁄x² + 2, and (x − 1⁄x)² = x² + 1⁄x² − 2.

Common traps

• Squaring x + 1⁄x as x² + 1⁄x², without the cross term 2.
• Writing cosec²A = cot²A − 1. The identity is cosec²A − cot²A = 1, so cosec²A = 1 + cot²A.

No directly related past PYQ was found.