SSC CGL 2025 Tier-I · 21 September 2025, Shift 1 · Quantitative Aptitude · Question 21 of 99

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 13 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 sin A = 4⁄5 and A is acute, find cot(90° − A).

(a)3⁄4
(b)4⁄3
(c)5⁄3
(d)5⁄4

Answer

Why

Correct — B.
Complementary-angle rule: cot(90° − A) = tan A
Find cos A: cos²A = 1 − sin²A = 1 − 16⁄25 = 9⁄25
A is acute, so take the positive root: cos A = 3⁄5

tan A = sin A ⁄ cos A = (4⁄5) ⁄ (3⁄5)
= 4⁄3 → option (b)

Why the others are wrong

(a)
3⁄4 — 3⁄4 is cot A = cos A ⁄ sin A. It is what you get by reading cot(90° − A) as cot A and skipping the switch to tan A.
(c)
5⁄3 — 5⁄3 is 1 ⁄ cos A = sec A (hypotenuse ⁄ adjacent in the 3-4-5 triangle), not tan A.
(d)
5⁄4 — 5⁄4 is 1 ⁄ sin A = cosec A (hypotenuse ⁄ opposite), not tan A.

Concept

A and 90° − A are complementary angles, and each ratio of one equals the co-ratio of the other: sin(90° − A) = cos A, tan(90° − A) = cot A, and cot(90° − A) = tan A.

With sin A = 4⁄5, the angle sits in a 3-4-5 right triangle: opposite 4, hypotenuse 5, so the adjacent side is 3. Every ratio then reads off the sides, and tan A = 4⁄3.

The word acute fixes the sign. Without it, cos A could be −3⁄5 and tan A could be −4⁄3.

Key facts

cot(90° − A) = tan A and tan(90° − A) = cot A
sin(90° − A) = cos A and cos(90° − A) = sin A
sin A = 4⁄5 with A acute gives cos A = 3⁄5, tan A = 4⁄3, sec A = 5⁄3 and cosec A = 5⁄4

Study next

sec(90° − A) = cosec A and cosec(90° − A) = sec AProducts such as tan 1° × tan 2° × … × tan 89° = 1, which pair A with 90° − AReading all six ratios from one right-triangle triple

Common traps

Treating cot(90° − A) as cot A and marking 3⁄4.
Mixing up the sides: sin A = 4⁄5 puts 4 opposite A, so tan A (opposite ⁄ adjacent) is 4⁄3, not 3⁄4.

The same complementary switch is set at 12 Sep 2025, 12:30, Quant Q.20: tan(90° − A) = √3 becomes cot A = √3, so A = 30° and sin A = 1⁄2.

Related PYQs

No directly related past PYQ was found.

Topic: Trigonometric ratio, Degree and Radian Measures, Standard Identities, Complementary angles, Heights and Distances — all 131 SSC CGL questions on it, every solved shift.