SSC CGL 2024 Tier-I · 9 September 2024, Shift 2 · Quantitative Aptitude · Question 12 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 5 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

Simplify the following. (sin³A − cos³A) ⁄ (sin A − cos A), where A is an acute angle.

(a)3 cosA - 1
(b)1 – 3 sinA
(c)sinA + cosA
(d)1 + sinA cosA

Answer

Why

Correct — D. The stem is printed as an image: simplify (sin³A − cos³A) ⁄ (sinA − cosA), where A is acute.

The numerator is a difference of cubes: a³ − b³ = (a − b)(a² + ab + b²), with a = sinA and b = cosA.

(sinA − cosA) cancels top and bottom
= sin²A + sinA cosA + cos²A
sin²A + cos²A = 1
= 1 + sinA cosA → option (d).

Check at A = 30°: the fraction is ≈ 1.433, and 1 + (1⁄2)(√3⁄2) ≈ 1.433.

Why the others are wrong

(a)
3 cosA - 1 — No cube factorisation leaves a term in cosA alone, so 3 cosA − 1 has no route in. At A = 30° it is ≈ 1.598 against the expression's ≈ 1.433.
(b)
1 – 3 sinA — 1 – 3 sinA is negative for most acute A — it is −0.5 at A = 30° — while 1 + sinA cosA always exceeds 1 for acute A. A sign check alone kills it.
(c)
sinA + cosA — sinA + cosA is what cancelling gives if you misread the numerator as the difference of squares sin²A − cos²A. It drops the middle term ab: 1.366 at A = 30°, not 1.433.

Concept

Three algebraic identities carry most of SSC's trigonometric simplification: a³ − b³, a³ + b³ and a² − b².

Dividing a³ − b³ by (a − b) leaves a² + ab + b². With a = sinA and b = cosA, the Pythagorean identity turns two of those three terms into 1, so only the product term survives.

Learn the companion result at the same time: (sin³A + cos³A) ⁄ (sinA + cosA) = 1 − sinA cosA. Same structure, opposite sign on the product.

The stem restricts A to acute angles, which keeps every term positive. The cancellation also needs sinA ≠ cosA — at A = 45° the original fraction is 0⁄0.

Key facts

a³ − b³ = (a − b)(a² + ab + b²), so (a³ − b³) ÷ (a − b) = a² + ab + b².
sin²A + cos²A = 1 for every angle A.
(sin³A + cos³A) ⁄ (sinA + cosA) = 1 − sinA cosA.

Study next

a³ + b³ and a² − b² applied to sin and cosVerifying a trigonometric identity by substituting A = 30°, 45° or 60°Expressions in sinA + cosA, using (sinA + cosA)² = 1 + 2 sinA cosA

Common traps

Cancelling as though the numerator were a difference of squares, which lands on option (c)
Dropping the middle term ab and answering 1
Writing the plus-cube result 1 − sinA cosA for this minus-cube expression

This shift leans hard on identity-spotting. Quant Q.3 (09 Sep 2024, 12:30) is the same trick in pure algebra — (a³ + b³ + c³ − 3abc) ⁄ (a² + b² + c² − ab − bc − ca) — and Quant Q.1 asks for sec t from tan t on the Pythagorean identity. The mark goes to whoever spots the factorisation instead of expanding.

Related PYQs

No directly related past PYQ was found.