SSC CGL 2024 Tier-I · 23 September 2024, Shift 3 · Quantitative Aptitude · Question 6 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 10 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.

If tan A = 1⁄√10, A is an acute angle, then the value of sin A + cosec A is:

Answer

Why

Correct — C. The stem, printed as an image, gives tan A = 1⁄√10 with A acute, and asks for sin A + cosec A.

tan A = opposite ⁄ adjacent, so take opposite = 1 and adjacent = √10.

Hypotenuse by Pythagoras:
h = √(1² + (√10)²) = √(1 + 10) = √11

So sin A = 1⁄√11 and cosec A = √11.
sin A + cosec A = 1⁄√11 + √11
= (1 + 11)⁄√11 = 12⁄√11 → option (c), which shows 12⁄√11.

Why the others are wrong

(a)
Option (a) shows 11⁄√10, built on the adjacent side √10 as if it were the hypotenuse. Both sin A and cosec A are measured against the hypotenuse, so √11 has to be the denominator.
(b)
Option (b) shows 10⁄√11. The denominator is right, the numerator is not: putting 1⁄√11 + √11 over the single denominator √11 gives 1 + (√11 × √11) = 12, not 10.
(d)
Option (d) shows 1⁄√10, which is tan A copied straight back from the stem. Nothing in sin A + cosec A reduces to the tangent.

Concept

One trigonometric ratio fixes the whole right triangle up to scale. From tan A = 1⁄√10 you may set the legs to 1 and √10 and build the hypotenuse — √11 — from Pythagoras.

Because A is acute, every ratio is positive and no sign case has to be argued.

The second half is an algebraic reflex, not a trigonometric one: sin A and cosec A are reciprocals, so if sin A = 1⁄h then their sum is 1⁄h + h = (1 + h²)⁄h. With h = √11 that is 12⁄√11.

12⁄√11 and its rationalised form 12√11⁄11 are the same number. Check which of the two forms the option list prints before you rationalise anything.

Key facts

tan A = opposite ⁄ adjacent, so tan A = 1⁄√10 fixes the two legs as 1 and √10.
The hypotenuse follows from Pythagoras as √(1 + 10) = √11.
sin A and cosec A are reciprocals, so their sum is (1 + h²)⁄h whenever sin A = 1⁄h.

Study next

Building the third side from any single ratio, then reading every other ratio off itUsing 1 + tan²A = sec²A as a shortcut to the hypotenuse without drawing the triangleSimplifying x + 1⁄x forms that arise from reciprocal trigonometric pairs

Common traps

Taking √10 as the hypotenuse because it is the larger number printed in tan A = 1⁄√10.
Adding 1⁄√11 and √11 as 1⁄√11 + 1⁄√11 by mistaking cosec A for a second sine.

One ratio is handed over and a combination of two others is demanded. A right-triangle value from tan A = 1 is set at 17 Sep 2024, 09:00, Quant Q.22, and a pure identity simplification of (cosec θ − sin θ)(sec θ − cos θ)(tan θ + cot θ) at 25 Sep 2024, 09:00, Quant Q.1.

Related PYQs

No directly related past PYQ was found.

Topic: Trigonometric ratio, Degree and Radian Measures, Standard Identities, Complementary angles, Heights and Distances — all 86 SSC CGL questions on it, every solved shift.