SSC CGL 2025 Tier-I · 19 September 2025, Shift 1 · 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.

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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 tanA + cotA = 2 and A ∈ (0, π⁄2), then what is sinA + cosA?

(a)1
(b)√2
(c)1⁄√2
(d)Cannot be determined

Answer

Why

Correct — B.
Combine the fractions: tanA + cotA = sinA⁄cosA + cosA⁄sinA = (sin²A + cos²A)⁄(sinA cosA) = 1⁄(sinA cosA)
Set it equal to 2: sinA cosA = 1⁄2

Square the target: (sinA + cosA)² = sin²A + cos²A + 2 sinA cosA
= 1 + 2 × 1⁄2 = 2
A lies in (0, π⁄2), where sine and cosine are positive, so take the positive root: sinA + cosA = √2 → option (b).

Why the others are wrong

(a)
1 — 1 is sin²A + cos²A, the identity, not the sum asked for. On [0, π⁄2], sinA + cosA = 1 only at the endpoints 0 and π⁄2, which the open interval excludes.
(c)
1⁄√2 — 1⁄√2 is sin 45° alone, or cos 45° alone. The question adds the two: 1⁄√2 + 1⁄√2 = 2⁄√2 = √2.
(d)
Cannot be determined — It can be determined. The equation fixes sinA cosA = 1⁄2 and the interval fixes the sign. Without the interval, A = 5π⁄4 would also fit and give −√2.

Concept

tanA + cotA = 1⁄(sinA cosA), because sin²A + cos²A = 1 sits in the numerator once the two fractions are combined.

That turns the given equation into a value for sinA cosA, and (sinA + cosA)² = 1 + 2 sinA cosA turns that into the sum. Squaring hides the sign, so the quadrant decides it.

A direct check: tanA + 1⁄tanA = 2 rearranges to (tanA − 1)² = 0, so tanA = 1 and A = π⁄4. Then sin 45° + cos 45° = 1⁄√2 + 1⁄√2 = √2, the same answer.

Key facts

tanA + cotA = 1⁄(sinA cosA) = secA cosecA.
(sinA + cosA)² = 1 + 2 sinA cosA.
(sinA − cosA)² = 1 − 2 sinA cosA.
For real x, x + 1⁄x = 2 holds only at x = 1, because it rearranges to (x − 1)² = 0.

Study next

(sinA − cosA)² and choosing the sign by quadrantsin⁴A + cos⁴A from a known value of sinA cosAConditions of the form x + 1⁄x = 2, such as sinA + cosecA = 2

Common traps

Taking the square root without using the interval to fix the sign
Answering sin²A + cos²A = 1 instead of sinA + cosA

25 Sep 2024, 09:00, Quant Q.1 leans on the same identity: (cosec θ − sin θ)(sec θ − cos θ)(tan θ + cot θ) reduces to sinθ cosθ × 1⁄(sinθ cosθ), keyed 1.

17 Sep 2025, 16:00, Quant Q.14 squares sinA + cosA with A in the second quadrant, where cosA = −4⁄5, keyed 1⁄25.

Related PYQs

No directly related past PYQ was found.

Topic: Trigonometric ratio, Degree and Radian Measures, Standard Identities, Complementary angles, Heights and Distances — all 98 SSC CGL questions on it, every solved shift.