SSC CGL 2025 Tier-I · 23 September 2025, Shift 3 · Quantitative Aptitude · Question 17 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 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 sinA = x, then what is the value of tanA in terms of x?

(a)x⁄√(1−x²)
(b)√(1−x²)⁄x
(c)1⁄x
(d)x²

Answer

Why

Correct — A. Find cos A from the identity, then divide.

cos²A = 1 − sin²A = 1 − x²
cos A = √(1 − x²), taking A acute so cos A is positive
tan A = sin A ÷ cos A = x⁄√(1 − x²) → option (a)

Triangle check: opposite side x, hypotenuse 1, so the adjacent side is √(1 − x²), and tan A = opposite ÷ adjacent.

Why the others are wrong

(b)
√(1−x²)⁄x — √(1−x²)⁄x is cos A ÷ sin A, which is cot A, the reciprocal of the ratio asked for.
(c)
1⁄x — 1⁄x is 1 ÷ sin A, which is cosec A, not tan A.
(d)
x² — x² is sin²A. Squaring sin A never brings in cos A, and tan A needs cos A in the denominator.

Concept

Any trigonometric ratio of an acute angle can be written from one given ratio using sin²A + cos²A = 1.

Draw a right triangle with hypotenuse 1 and opposite side x. Pythagoras gives the adjacent side √(1 − x²), and every other ratio reads straight off the triangle.

When sin A is a fraction p⁄q, take the opposite side as p and the hypotenuse as q.

The stem does not say A is acute. If cos A were negative, tan A would be −x⁄√(1 − x²), which is not among the options, so the key takes the positive root. For tan A to exist, x must also lie strictly between −1 and 1.

Key facts

sin²A + cos²A = 1, so cos A = √(1 − sin²A) for acute A.
tan A = sin A ÷ cos A, and cot A = cos A ÷ sin A.
cosec A = 1 ÷ sin A and sec A = 1 ÷ cos A.

Study next

Writing all six ratios from one given ratioStandard values of sin, cos and tan at 0°, 30°, 45°, 60° and 90°Identities sec²A − tan²A = 1 and cosec²A − cot²A = 1

Common traps

Inverting the ratio: tan A is opposite over adjacent, and √(1 − x²)⁄x is cot A.
Dropping the root: x⁄(1 − x²) divides by cos²A instead of cos A.

The first step alone is 19 Sep 2025, 16:00, Quant Q.22: from sin A = x, cos²A = 1 − x².

A disguised version is 24 Sep 2025, 12:30, Quant Q.20: cos(90° − θ) = x is sin θ = x, so cos θ = √(1 − x²).

Related PYQs

No directly related past PYQ was found.

Topic: Basic algebraic identities of School Algebra & Elementary surds — all 182 SSC CGL questions on it, every solved shift.