SSC CGL 2025 Tier-I · 21 September 2025, Shift 3 · Quantitative Aptitude · Question 18 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.

IfsinA = 1⁄√10, then find the value oftanA + secA.

(a)(1+√10)⁄3
(b)(5+√10)⁄5
(c)(6+√10)⁄3
(d)(1+√10)⁄5

Answer

Why

Correct — A. sin A = opposite ⁄ hypotenuse, so draw a right triangle with opposite 1 and hypotenuse √10.

Pythagoras: adjacent = √(10 − 1) = √9 = 3
tan A = opposite ⁄ adjacent = 1⁄3
sec A = hypotenuse ⁄ adjacent = √10⁄3

Add over the common denominator 3: 1⁄3 + √10⁄3 = (1 + √10)⁄3 → option (a)

Why the others are wrong

(b)
(5+√10)⁄5 — (5 + √10)⁄5 has the wrong denominator. Both tan A = 1⁄3 and sec A = √10⁄3 sit over 3, so their sum does too. It is about 1.63 against the correct 1.39.
(c)
(6+√10)⁄3 — (6 + √10)⁄3 splits as 2 + √10⁄3. The sec A part is right, but it needs tan A = 2, while tan A = opposite ⁄ adjacent = 1⁄3.
(d)
(1+√10)⁄5 — (1 + √10)⁄5 has the right numerator over the wrong denominator. Both ratios divide by the adjacent side, √(10 − 1) = 3, and no side of this triangle measures 5.

Concept

Given one trigonometric ratio, build the right triangle it describes and read every other ratio off its three sides.

sin A = 1⁄√10 fixes opposite 1 and hypotenuse √10. Pythagoras gives the third side: adjacent = √(10 − 1) = 3. So tan A = 1⁄3, cos A = 3⁄√10 and sec A = √10⁄3.

A second route skips the triangle: tan A + sec A = (1 + sin A)⁄cos A. With cos A = 3⁄√10 this is (1 + 1⁄√10) × √10⁄3 = (√10 + 1)⁄3, the same answer.

The stem never says A is acute. In the second quadrant cos A = −3⁄√10 and the sum is −(1 + √10)⁄3, which no option offers, so the options assume the acute angle.

Key facts

sin A = 1⁄√10 fixes a right triangle with sides 1, 3 and √10.
tan A = opposite ⁄ adjacent and sec A = hypotenuse ⁄ adjacent = 1 ⁄ cos A.
tan A + sec A = (1 + sin A) ⁄ cos A.

Study next

sec A − tan A from sec A + tan A, using sec²A − tan²A = 1Signs of the ratios when the angle lies in the second quadrant1 + tan²A written in terms of sin A

Common traps

Dropping the square root in Pythagoras: the adjacent side is √9 = 3, not 9.
Taking sec A as 1 ⁄ sin A, which is cosec A (here √10), instead of 1 ⁄ cos A.

19 Sep 2025, 09:00, Quant Q.22 gives sin x = 3⁄5 and asks for (1 + tan x)⁄(1 − tan x). The 3-4-5 triangle gives tan x = 3⁄4, keyed 7.

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

Related PYQs

No directly related past PYQ was found.

Topic: Square roots — all 21 SSC CGL questions on it, every solved shift.