SSC CGL 2024 Tier-I · 11 September 2024, Shift 3 · Quantitative Aptitude · Question 2 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 6 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.

If p sin A − cos A = 1, then p² − (1 + p²) cos A equals:

(a)0
(b)1
(c)-1
(d)2

Answer

Why

Correct — B. Isolate p, then write everything in cos A.

p sin A − cos A = 1 gives p sin A = 1 + cos A
p = (1 + cos A)⁄sin A

Since sin²A = 1 − cos²A = (1 − cos A)(1 + cos A),
p² = (1 + cos A)²⁄sin²A = (1 + cos A)⁄(1 − cos A)

1 + p² = [(1 − cos A) + (1 + cos A)]⁄(1 − cos A) = 2⁄(1 − cos A)

p² − (1 + p²)cos A = (1 + cos A − 2cos A)⁄(1 − cos A) = (1 − cos A)⁄(1 − cos A) = 1 → option (b).

Why the others are wrong

(a)
0 — 0 would need the numerator 1 − cos A to vanish, that is cos A = 1 and A = 0°. But then sin A = 0 and p is undefined, so the expression never takes the value 0.
(c)
-1 — −1 is the last subtraction taken the wrong way round. Computing (1 + p²)cos A − p² in place of p² − (1 + p²)cos A gives (cos A − 1)⁄(1 − cos A), which is −1.
(d)
2 — 2 is the numerator of 1 + p², which is 2⁄(1 − cos A). Reading that fraction as the answer skips the last two operations, multiplying by cos A and subtracting.

Concept

Everything here rests on sin²A + cos²A = 1, used in its factorised form: 1 − cos²A = (1 − cos A)(1 + cos A).

That factorisation is what lets (1 + cos A)² over sin²A collapse to a single ratio, (1 + cos A)⁄(1 − cos A), with no square roots left.

The result does not depend on A at all. Because the expression is constant in the angle, the question has a single number as its answer whatever A you take — which is what makes the check below legitimate.

Because the value is independent of A, a legal shortcut is to pick a convenient angle. At A = 60°: p(√3⁄2) − 1⁄2 = 1 gives p = √3, so p² = 3, and 3 − (1 + 3) × 1⁄2 = 3 − 2 = 1.

Key facts

sin²A + cos²A = 1, so 1 − cos²A factorises as (1 − cos A)(1 + cos A).
From p sin A − cos A = 1 you get p = (1 + cos A)⁄sin A.
p² = (1 + cos A)⁄(1 − cos A) and 1 + p² = 2⁄(1 − cos A).
Testing A = 60° gives p = √3 and the expression evaluates to 1.

Study next

Factorising 1 − cos²A and 1 − sin²A to clear a trigonometric denominatorExpressions in sin and cos whose value is independent of the angleSubstituting a convenient angle to test a trigonometric identity

Common traps

Squaring term by term to p²sin²A − cos²A = 1, which is not what squaring a difference gives.
Cancelling (1 − cos A) against itself and writing 0 instead of 1.
Abandoning the algebra because the answer looks as though it should depend on A.

Quant Q.7 of this same shift asks you to rewrite sin 74° + tan 74° using complementary angles — the same subject with none of the algebra.

Look first for the identity that removes a squared term. Here it is sin²A = 1 − cos²A, and once that is used the expression has one variable and cancels.

Related PYQs

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