SSC CGL 2025 Tier-I · 18 September 2025, Shift 2 · Quantitative Aptitude · Question 22 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 = 2, then sinA−cosA = ?

(a)2⁄√5
(b)−1⁄√5
(c)0
(d)3⁄5

Answer

Why

Correct — B. Build a right triangle from cot A = adjacent ⁄ opposite = 2⁄1, taking A as acute.

Adjacent side = 2, opposite side = 1
Hypotenuse by Pythagoras: √(2² + 1²) = √5
Read off the ratios: sin A = 1⁄√5, cos A = 2⁄√5

Subtract: sin A − cos A = 1⁄√5 − 2⁄√5 = −1⁄√5 → option (b)

Why the others are wrong

(a)
2⁄√5 — 2⁄√5 is cos A on its own, not the difference. Subtracting it from sin A = 1⁄√5 leaves −1⁄√5.
(c)
0 — sin A − cos A = 0 needs sin A = cos A, which means cot A = 1 and A = 45°. With cot A = 2, cos A is twice sin A.
(d)
3⁄5 — 3⁄5 is cos²A − sin²A (4⁄5 − 1⁄5), a difference of squares. sin A − cos A itself is negative, because cos A is larger than sin A.

Concept

One trigonometric ratio fixes a right triangle up to scale. cot A = adjacent ⁄ opposite, so cot A = 2 gives sides 2 and 1, and Pythagoras gives the hypotenuse √5.

Every other ratio is then read from that triangle. Since cot A > 1, the adjacent side is longer than the opposite, so cos A > sin A and sin A − cos A must be negative. −1⁄√5 is the one negative value among the four choices.

The question does not say A is acute. cot A = 2 also holds in the third quadrant, where sin A = −1⁄√5 and cos A = −2⁄√5 give sin A − cos A = +1⁄√5, which is not among the choices. The key takes A as acute.

Key facts

cot A = cos A ⁄ sin A = adjacent ⁄ opposite.
cot A = 2 with A acute gives sin A = 1⁄√5 and cos A = 2⁄√5.
cosec²A = 1 + cot²A, so cot A = 2 gives cosec²A = 5, a check on sin A = 1⁄√5.
For acute A, cos A > sin A exactly when A < 45°, that is, when cot A > 1.

Study next

Finding all six ratios from one given ratio with a right triangleThe identities cosec²A = 1 + cot²A and sec²A = 1 + tan²ASigns of the trigonometric ratios in each quadrant

Common traps

Reading cot as opposite ⁄ adjacent: that is tan, and it swaps sin and cos, giving +1⁄√5.
Stopping at cos A = 2⁄√5 and forgetting to subtract it from sin A.

21 Sep 2025, 16:00, Quant Q.18 uses the same triangle method from sin A = 1⁄√10: sides 1, 3 and √10 give tan A + sec A = 1⁄3 + √10⁄3 = (1 + √10)⁄3.

17 Sep 2025, 16:00, Quant Q.22 also starts from cot A and uses cosec²A = 1 + cot²A: cot A = x + 1⁄x gives cosec²A = x² + 1⁄x² + 3.

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.