SSC CGL 2024 Tier-I · 13 September 2024, Shift 3 · Quantitative Aptitude · Question 20 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 10 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 m = a secA and y = b tanA, then find the value of b²m² − a²y² + a²y² ⁄ b²m² + cos²A.

(a)1-a² b²
(b)a² b²
(c)a² b²+1
(d)a² b²+2

Answer

Why

Correct — C. The stem is printed as an image and reads: if m = a secA and y = b tanA, then find the value of b²m² − a²y² + a²y²⁄(b²m²) + cos²A.

Square and substitute:
b²m² = a²b² sec²A and a²y² = a²b² tan²A

First pair: b²m² − a²y² = a²b²(sec²A − tan²A) = a²b²

Fraction: a²y²⁄(b²m²) = tan²A⁄sec²A = sin²A
Add cos²A: sin²A + cos²A = 1

Total = a²b² + 1 → option (c)

Why the others are wrong

(a)
1-a² b² — The sign of the first pair is flipped. b²m² − a²y² = a²b²(sec²A − tan²A), and sec²A − tan²A is +1, so a²b² enters with a plus.
(b)
a² b² — a²b² stops after the first pair. The fraction and cos²A still add sin²A + cos²A = 1 to the total.
(d)
a² b²+2 — + 2 needs the fraction to be 1, but a²y²⁄(b²m²) = tan²A⁄sec²A = sin²A. With cos²A that tail is exactly 1.

Concept

When a stem defines letters as multiples of sec and tan, square and substitute. The constants a and b ride along as factors and the angle drops out through an identity.

sec²A − tan²A = 1 clears the first pair. Dividing tan²A by sec²A leaves sin²A, and the stem's own cos²A completes sin²A + cos²A = 1.

Nothing needs solving — every piece is built to collapse.

The options print the product as a² b², with a space, so option (c) reads a²b² + 1.

A spot check at A = 0°: m = a and y = 0, so the expression is a²b² − 0 + 0 + cos²0° = a²b² + 1.

Key facts

sec²A − tan²A = 1 wherever both are defined.
tan²A ⁄ sec²A = sin²A, because tan A ⁄ sec A = sin A.
With m = a sec A and y = b tan A, b²m² − a²y² = a²b².

Study next

Eliminating θ from x = a sec θ, y = b tan θ to reach x²⁄a² − y²⁄b² = 1Using sec θ − tan θ = 1⁄(sec θ + tan θ) to split a given sumRewriting tan, sec and cosec in sin and cos before simplifying

Common traps

Taking sec²A − tan²A as −1, which is the value of tan²A − sec²A
Cancelling a²y²⁄(b²m²) to 1 because the letters look symmetric
Stopping at a²b² and forgetting that the fraction and cos²A still add 1

The identity sec²θ − tan²θ = 1 also carries 25 Sep 2024, 16:00, Quant Q.14: from sec θ + tan θ = x it gives sec θ − tan θ = 1⁄x, and sin θ comes out as (x² − 1)⁄(1 + x²).

Related PYQs

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Topic: Basic algebraic identities of School Algebra & Elementary surds — all 130 SSC CGL questions on it, every solved shift.