The value of 1⁄(cosecθ − cotθ) − 1⁄sinθ is equal to:

Correct — A. The stem asks for the value of 1 ⁄ (cosec θ − cot θ) − 1 ⁄ sin θ.

Rule: cosec²θ − cot²θ = 1, so multiplying by the conjugate clears the first denominator outright.

1 ⁄ (cosec θ − cot θ)
= (cosec θ + cot θ) ⁄ [(cosec θ − cot θ)(cosec θ + cot θ)]
= (cosec θ + cot θ) ⁄ 1 = cosec θ + cot θ

Now the second term, since 1 ⁄ sin θ = cosec θ:
(cosec θ + cot θ) − cosec θ = cot θ → option (a), the image reading cot θ

Why the others are wrong

• (b)
  Option (b) shows sec θ. Cosine does enter, through cot θ = cos θ ⁄ sin θ, but nothing in the working ever divides by cos θ, so the 1 ⁄ cos θ that sec θ needs is never formed.
• (c)
  Option (c) shows cosec θ. That is what survives if you cancel the cot θ at the subtraction step, but the term 1 ⁄ sin θ removes the cosec θ, leaving cot θ.
• (d)
  Option (d) shows tan θ, the reciprocal of the right value. Test θ = 30°: the expression works out to √3, which is cot 30°, while tan 30° is 1 ⁄ √3.

Three Pythagorean identities carry most of SSC trigonometry: sin² + cos² = 1, 1 + tan² = sec², and 1 + cot² = cosec².

Rearrange the last one and you get cosec²θ − cot²θ = 1 — a difference of squares equal to exactly 1. That is why the conjugate trick works so cleanly here.

Whenever a denominator has the shape (cosec θ − cot θ) or (sec θ − tan θ), multiply top and bottom by its conjugate. The denominator becomes 1 and the fraction vanishes, leaving a sum you can finish in your head.

Key facts

• cosec²θ − cot²θ = 1, which is 1 + cot²θ = cosec²θ rearranged.
• Therefore 1 ÷ (cosec θ − cot θ) = cosec θ + cot θ.
• The same conjugate move gives 1 ÷ (sec θ − tan θ) = sec θ + tan θ.
• 1 ÷ sin θ is cosec θ, which is why the cosec terms cancel and cot θ is left.

Common traps

• Rationalising correctly and then forgetting to subtract 1 ÷ sin θ, leaving cosec θ + cot θ
• Inverting term by term and writing 1 ÷ (cosec θ − cot θ) as sin θ − tan θ
• Cancelling the cot θ rather than the cosec θ at the final subtraction

No directly related past PYQ was found.