The value of is:

Correct — B. Factor the top and the bottom of the bracket — a cos 2θ falls out of both.

sin θ − 2sin³θ = sin θ(1 − 2sin²θ) = sin θ · cos 2θ
2cos³θ − cos θ = cos θ(2cos²θ − 1) = cos θ · cos 2θ
The cos 2θ cancels, leaving sin θ ⁄ cos θ = tan θ

So the bracket cubed is tan³θ, and tan³θ × (1 ⁄ tan θ) = tan²θ.

tan²θ − sec²θ = −(sec²θ − tan²θ) = −1 → option (b).

Why the others are wrong

• (a)
  2 — 2 is a coefficient printed inside the bracket, not a value the expression can take. sec²θ exceeds tan²θ by exactly 1, so their difference is fixed whatever θ is.
• (c)
  1 — 1 is sec²θ − tan²θ, the identity taken in the opposite order. This expression subtracts sec²θ from tan²θ, so the sign flips.
• (d)
  0 — 0 assumes sec²θ = tan²θ, dropping the 1 in sec²θ = 1 + tan²θ. That 1 is the whole answer here.

Two double-angle forms are hiding in the bracket: 1 − 2sin²θ = cos 2θ and 2cos²θ − 1 = cos 2θ. They are the same quantity, so once both are factored out the bracket collapses to tan θ.

After that the item is a single identity, sec²θ − tan²θ = 1. The cube and the 1 ⁄ tan θ exist only to turn tan θ into tan²θ, so the final subtraction is forced.

The value does not depend on θ at all. That independence is the signal that an identity, not an angle, is being tested.

Key facts

• cos 2θ = 1 − 2sin²θ = 2cos²θ − 1.
• sec²θ − tan²θ = 1, so tan²θ − sec²θ = −1 wherever both are defined.
• (sin θ − 2sin³θ) ⁄ (2cos³θ − cos θ) simplifies to tan θ.

Common traps

• Substituting a convenient angle such as 45°, where this expression is undefined.
• Cancelling sin θ against cos θ rather than cos 2θ against cos 2θ.
• Applying the cube to the numerator of the bracket alone.

No directly related past PYQ was found.