Find the value of (cosec θ - sin θ )(sec θ - cos θ )(tan θ + cot θ ).

Correct — D. Turn each bracket into a single fraction first, then cancel.

cosec θ − sin θ = 1⁄sin θ − sin θ = (1 − sin²θ)⁄sin θ = cos²θ⁄sin θ
sec θ − cos θ = 1⁄cos θ − cos θ = (1 − cos²θ)⁄cos θ = sin²θ⁄cos θ
tan θ + cot θ = (sin²θ + cos²θ)⁄(sin θ cos θ) = 1⁄(sin θ cos θ)

Multiply the three:
(cos²θ⁄sin θ) × (sin²θ⁄cos θ) = sin θ cos θ
sin θ cos θ × 1⁄(sin θ cos θ) = 1

The four options are printed as pictures of numbers, and 1 is option (d).

Why the others are wrong

• (a)
  Option (a) is the picture of 1⁄2. Nothing in the product halves anything — the three brackets reduce to sin θ cos θ over sin θ cos θ, and that quotient is exactly one.
• (b)
  Option (b) is 0. A zero needs one factor to vanish, but cos²θ⁄sin θ, sin²θ⁄cos θ and 1⁄(sin θ cos θ) are all non-zero wherever the expression is defined.
• (c)
  Option (c) is −1. The sign of the product is sign(sin θ) × sign(cos θ) × sign(sin θ cos θ), which is positive for every θ, so a minus sign cannot appear.

The reliable first move on any identity of this shape is to write everything in sin and cos. Once you do, each bracket becomes a single fraction and the cancellation is visible.

The two Pythagorean rearrangements do the work: 1 − sin²θ = cos²θ and 1 − cos²θ = sin²θ.

The third bracket is the standard result tan θ + cot θ = 1⁄(sin θ cos θ), which is just sin²θ + cos²θ = 1 over a common denominator.

Key facts

• cosec θ − sin θ = cos²θ⁄sin θ, because 1 − sin²θ = cos²θ.
• sec θ − cos θ = sin²θ⁄cos θ, because 1 − cos²θ = sin²θ.
• tan θ + cot θ = 1⁄(sin θ cos θ), since sin²θ + cos²θ = 1.
• The whole product equals 1 for every θ at which it is defined.

Common traps

• Expanding the brackets term by term, which produces eight terms and no cancellation.
• Cancelling sin θ cos θ only once when it appears twice over in the product.
• Misreading the option pictures at small size, where 1 and −1 differ by one short dash.

No directly related past PYQ was found.