The value of (cosecA ⁄ (cosecA − 1) + cosecA ⁄ (cosecA + 1)) (secA ⁄ (secA − 1) + secA ⁄ (secA + 1))⁻¹, when A = 60° is:

Correct — A. Simplify each bracket before you substitute; the expression collapses to tan²A.

First bracket = cosecA[1⁄(cosecA − 1) + 1⁄(cosecA + 1)]
= 2cosec²A⁄(cosec²A − 1) = 2cosec²A⁄cot²A = 2sec²A

Second bracket = secA[1⁄(secA − 1) + 1⁄(secA + 1)]
= 2sec²A⁄(sec²A − 1) = 2sec²A⁄tan²A = 2cosec²A

The ⁻¹ inverts the second bracket, so the value is
2sec²A ÷ 2cosec²A = sin²A⁄cos²A = tan²A

tan 60° = √3, so tan²60° = 3 → option (a).

Why the others are wrong

• (b)
  4 — 4 is sec²60°. You land there by stopping one Pythagorean step short — sec²A = 1 + tan²A, so 4 is the answer plus one.
• (c)
  1 — 1 would need the two brackets to be equal. They are not: one is 2sec²A and the other 2cosec²A, and those coincide only at A = 45°.
• (d)
  2 — 2 is sec 60° itself, a value that appears mid-working. The expression reduces to tan²A, and squaring √3 carries it past 2 to 3.

Both brackets are the same algebraic shape: x⁄(x − 1) + x⁄(x + 1) = 2x²⁄(x² − 1).

That denominator is what makes the question work, because cosec²A − 1 = cot²A and sec²A − 1 = tan²A. Each bracket therefore turns into a single reciprocal square.

Everything up to the last line is angle-free: the expression equals tan²A for any A where it is defined. The 60° only arrives to be evaluated.

Key facts

• cosec²A − 1 = cot²A, one of the two Pythagorean identities this item is built on.
• sec²A − 1 = tan²A, the other one.
• x⁄(x − 1) + x⁄(x + 1) = 2x²⁄(x² − 1) for any x, which is why both brackets collapse alike.
• tan 60° = √3, so tan²60° = 3.

Common traps

• Substituting A = 60° first and grinding through four messy fractions
• Missing the ⁻¹ and multiplying the two brackets instead of dividing
• Swapping the identities, so cosec²A − 1 is written as tan²A

No directly related past PYQ was found.