SSC CGL 2025 Tier-I · 19 September 2025, Shift 3 · Quantitative Aptitude · Question 14 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 11 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

If cotA = √3, what is the value of(1 + sinA)(1 + cosA)?

(a)(2+3√3)⁄2
(b)(6+3√3)⁄4
(c)(7+3√3)⁄4
(d)(11+3√3)⁄5

Answer

Why

Correct — B. Read the angle from cot A, then substitute.

cot A = √3 = cot 30°, so A = 30°
sin A = 1⁄2, cos A = √3⁄2

Add 1 to each: 1 + sin A = 3⁄2 and 1 + cos A = (2 + √3)⁄2
Multiply: (3⁄2) × (2 + √3)⁄2 = 3(2 + √3)⁄4
= (6 + 3√3)⁄4 → option (b)

Why the others are wrong

(a)
(2+3√3)⁄2 — (2 + 3√3)⁄2 is (4 + 6√3)⁄4 over the same denominator. Both parts are off: the product's constant is 6⁄4 and its √3 term 3√3⁄4.
(c)
(7+3√3)⁄4 — (7 + 3√3)⁄4 is exactly 1⁄4 more than the true value. The constant part is 1 + sin A = 1 + 1⁄2 = 6⁄4, not 7⁄4.
(d)
(11+3√3)⁄5 — A denominator of 5 cannot arise: sin 30° and cos 30° both have denominator 2, so the product has 4. The true value is (6 + 3√3)⁄4 ≈ 2.80.

Concept

When a ratio equals a standard value, name the angle first. √3 is cot 30° (and tan 60°), so cot A = √3 fixes A = 30° for an acute angle.

For any other value, draw the right triangle: cot A = adjacent ⁄ opposite = √3 ⁄ 1, so the hypotenuse is √(3 + 1) = 2. That gives sin A = 1⁄2 and cos A = √3⁄2 without memorising anything.

The stem does not say A is acute. cot A is also √3 at A = 210°, where sin A = −1⁄2 and cos A = −√3⁄2, and the product becomes (2 − √3)⁄4, which matches no option. The acute reading is the one the options support.

Key facts

cot 30° = √3 and tan 30° = 1⁄√3.
sin 30° = 1⁄2 and cos 30° = √3⁄2.
(1 + sin A)(1 + cos A) = 1 + sin A + cos A + sin A cos A.

Study next

Finding sin and cos from a non-standard cot or tan through a right triangleStandard values of sin, cos, tan and cot at 30°, 45° and 60°Products such as (1 − sin A)(1 + sin A) = cos²A

Common traps

Reading cot A = √3 as A = 60°. It costs nothing here only because the expression is symmetric in sin A and cos A, and it will cost you in a sin A − cos A item.
Dropping the sin A cos A term when expanding, which leaves (6 + 2√3)⁄4

18 Sep 2025, 12:30, Quant Q.22 gives a non-standard cot A = 2 and asks for sin A − cos A: the triangle with sides 2, 1, √5 gives a keyed −1⁄√5.

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