SSC CGL 2025 Tier-I · 19 September 2025, Shift 1 · 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 sinx = 3⁄5 and x ∈ (0, π⁄2), then find(1+tan x)⁄(1−tan x)

(a)6
(b)7
(c)8
(d)9

Answer

Why

Correct — B.
sin x = 3⁄5, so take opposite = 3 and hypotenuse = 5
Adjacent side: √(5² − 3²) = √16 = 4
x lies in (0, π⁄2), so tan x = 3⁄4, positive

Numerator: 1 + 3⁄4 = 7⁄4
Denominator: 1 − 3⁄4 = 1⁄4
Divide: (7⁄4) ÷ (1⁄4) = 7 → option (b)

Why the others are wrong

(a)
6 — 6 would need tan x = 5⁄7, since (1 + t) ⁄ (1 − t) = 6 gives t = 5⁄7. The 3-4-5 triangle gives tan x = 3⁄4.
(c)
8 — 8 would need tan x = 7⁄9, from (1 + t) ⁄ (1 − t) = 8. No ratio of the sides 3, 4 and 5 equals 7⁄9.
(d)
9 — 9 comes from using 4⁄5: (1 + 4⁄5) ÷ (1 − 4⁄5) = 9. But 4⁄5 is cos x, and tan x = sin x ⁄ cos x = 3⁄4.

Concept

Given one trigonometric ratio of an acute angle, draw the right triangle. sin x = 3⁄5 means opposite 3 and hypotenuse 5, Pythagoras gives adjacent 4, and every other ratio can be read off.

The expression is also a known form: (1 + tan x) ⁄ (1 − tan x) = tan(45° + x), from tan(A + B) = (tan A + tan B) ⁄ (1 − tan A tan B) with tan 45° = 1.

The stem is printed as an image. It reads: if sin x = 3⁄5 and x ∈ (0, π⁄2), find (1 + tan x) ⁄ (1 − tan x).

The interval matters: it makes cos x = +4⁄5, not −4⁄5, and so tan x positive.

Key facts

For an acute angle, sin x = 3⁄5 gives cos x = 4⁄5 and tan x = 3⁄4.
tan x = sin x ⁄ cos x.
(1 + tan x) ⁄ (1 − tan x) = tan(45° + x).

Study next

Finding all six ratios from one given ratioSigns of sin, cos and tan in each quadrantThe tan(A + B) and tan(A − B) formulas

Common traps

Taking cos x = 4⁄5 as tan x, which leads to 9
Ignoring the interval, which fixes the signs of cos x and tan x

17 Sep 2025, 16:00, Quant Q.14 starts from the same sin A = 3⁄5 but puts A in the second quadrant, so cos A = −4⁄5 and (sin A + cos A)² = the keyed 1⁄25.

18 Sep 2025, 12:30, Quant Q.14 pairs sin A = 3⁄5 with cos B = 12⁄13, both acute: sin(A + B) = 36⁄65 + 20⁄65 = the keyed 56⁄65.

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.