If cot A = x + 1⁄x, find cosec²A.

Correct — C. Use the identity linking cosec and cot, then square the given expression.

Identity: cosec²A = 1 + cot²A
Square cot A: (x + 1⁄x)² = x² + 2 · x · 1⁄x + 1⁄x²
Simplify the cross term, x · 1⁄x = 1: cot²A = x² + 1⁄x² + 2
Add 1: cosec²A = x² + 1⁄x² + 3 → option (c)

Why the others are wrong

• (a)
  x² + 1⁄x² + 1 — x² + 1⁄x² + 1 squares x + 1⁄x as x² + 1⁄x², dropping the cross term 2, then adds 1. The square of a sum always carries 2 · x · 1⁄x = 2.
• (b)
  x² + 1⁄x² + 2 — x² + 1⁄x² + 2 is cot²A, not cosec²A. It stops one step short: the identity adds 1 to cot²A.
• (d)
  x² + 1⁄x² + 4 — x² + 1⁄x² + 4 adds 2 to cot²A. But cosec²A − cot²A = 1, so cosec²A sits exactly 1 above cot²A = x² + 1⁄x² + 2.

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

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

Key facts

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

Common traps

• Writing (x + 1⁄x)² as x² + 1⁄x² and losing the cross term 2.
• Stopping at cot²A and forgetting the identity's + 1.
• Reversing the identity to cosec²A = cot²A − 1.

No directly related past PYQ was found.