SSC CGL 2024 Tier-I · 9 September 2024, Shift 2 · Quantitative Aptitude · Question 1 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 5 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

What is the value of sec (t), if tan(t) = 1⁄3?

Answer

Why

Correct — C. The stem gives tan t = 1⁄3 and asks for sec t, so use the identity that bridges those two: sec²t = 1 + tan²t.

sec²t = 1 + tan²t
= 1 + (1⁄3)² = 1 + 1⁄9
= 10⁄9

Square-root the top and the bottom:
sec t = √10⁄√9
= √10⁄3 → option (c)

The triangle route agrees. Label opposite = 1 and adjacent = 3, so the hypotenuse is √(1² + 3²) = √10, and sec t = hypotenuse⁄adjacent = √10⁄3.

Why the others are wrong

(a)
Below 1, so it cannot be a secant at all. |sec t| ≥ 1 always, because sec t = 1⁄cos t and |cos t| ≤ 1. √3⁄3 ≈ 0.58 is the value of tan 30°, which nothing here asks for.
(b)
Ruled out by the same test — 2√2⁄3 = √(8⁄9) ≈ 0.94, again below 1. It is what you get by reading the given 1⁄3 as sin t and computing cos t = √(1 − 1⁄9).
(d)
Right numerator, unfinished denominator. Stopping at √10⁄9 ≈ 0.35 roots only the top and lands below 1 again — both parts must be rooted: √10⁄√9 = √10⁄3.

Concept

Once any one ratio of an acute angle is known, every other ratio is fixed. The shortest bridge from tan to sec is 1 + tan²t = sec²t, which is just sin²t + cos²t = 1 divided through by cos²t.

The triangle picture makes it concrete. tan t = opposite⁄adjacent = 1⁄3 lets you label the two legs 1 and 3, and Pythagoras fixes the hypotenuse at √10.

Every remaining ratio then reads straight off that same triangle: sec t = √10⁄3, cos t = 3⁄√10, sin t = 1⁄√10, cosec t = √10.

Strictly, tan t = 1⁄3 also holds in the third quadrant, where sec t = −√10⁄3.

All four options here are positive, so the intended t is acute. Read SSC trigonometry items as first-quadrant unless the paper states a quadrant or a range.

Key facts

1 + tan²t = sec²t, obtained by dividing sin²t + cos²t = 1 through by cos²t.
1 + cot²t = cosec²t, the same identity divided by sin²t instead.
|sec t| ≥ 1 and |cosec t| ≥ 1 always, so any option strictly between −1 and 1 can be struck out before calculating.
For acute t with tan t = p⁄q, the hypotenuse is √(p² + q²), so sec t = √(p² + q²)⁄q.

Study next

Reading cos, sin, cot and cosec off a single given ratio using the 1–3–√10 style triangleChoosing between sin²+cos²=1, 1+tan²=sec² and 1+cot²=cosec² by which two functions the question namesValues of tan, sec and cosec at 0°, 30°, 45°, 60° and 90°, including where each is undefined

Common traps

Square-rooting only the numerator of sec²t = 10⁄9 and answering √10⁄9.
Reading the given 1⁄3 as sin t or cos t rather than tan t.
Rationalising out of habit — √10⁄3 already has a rational denominator.

SSC hands you one ratio and asks for another, choosing numbers that make the third side a surd such as √10 or √13 so the answer looks harder than the work is. The same stem returns as 'find sec t + tan t' or 'sec t − tan t', which the identical triangle answers.

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