SSC CGL 2024 Tier-I · 25 September 2024, Shift 1 · Quantitative Aptitude · Question 24 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 m + 5 = sec²A cosec²A(1 − cos²A)(1 − sin²A) + 5, then find the value of m:

(a)0
(b)5
(c)1
(d)10

Answer

Why

Correct — C. Rewrite the two brackets with the Pythagorean identity before multiplying anything out.

1 − cos²A = sin²A
1 − sin²A = cos²A

The right-hand side is now sec²A · cosec²A · sin²A · cos²A + 5.

sec²A = 1⁄cos²A and cosec²A = 1⁄sin²A, so the four factors cancel in pairs:
(1⁄cos²A)(1⁄sin²A)(sin²A)(cos²A) = 1

m + 5 = 1 + 5 = 6, so m = 1 → option (c).

Why the others are wrong

(a)
0 — 0 is what expanding (1 − cos²A)(1 − sin²A) as 1 − sin²A − cos²A gives, since that is 1 − 1. The expansion also carries a + sin²A cos²A term, and the product is sin²A cos²A, not zero.
(b)
5 — 5 is the constant sitting on both sides, which simply cancels. Choosing it means reading the right-hand side as 5 and missing the 1 the trigonometric product contributes.
(d)
10 — 10 treats the equation as m + 5 = 5 + 5, that is, the trigonometric block as 5. That block is sec²A cosec²A sin²A cos²A, which is 1 wherever the functions are defined.

Concept

The whole item is two substitutions and a cancellation.

sin²A + cos²A = 1 rewrites each bracket: 1 − cos²A becomes sin²A, and 1 − sin²A becomes cos²A.

sec A = 1⁄cos A and cosec A = 1⁄sin A turn the two squared reciprocals into denominators — and the brackets you have just rewritten are exactly those denominators.

Every factor pairs off, the product is 1, the +5 on each side cancels, and m = 1. The result holds for every A where sin A and cos A are both non-zero, so no particular angle is involved.

The +5 on both sides changes no mathematics, but it is what separates m from the value of the whole right-hand side, which is 6.

Substituting a friendly angle is a fair check: at A = 45°, sec²A cosec²A = 2 × 2 = 4 and sin²A cos²A = ½ × ½ = ¼, so the product is 4 × ¼ = 1.

Key facts

sin²A + cos²A = 1, so 1 − cos²A = sin²A and 1 − sin²A = cos²A.
sec²A = 1⁄cos²A and cosec²A = 1⁄sin²A.
sec²A · cosec²A · sin²A · cos²A = 1 wherever sin A and cos A are both non-zero.
Here the right-hand side comes to 6, so m = 1.

Study next

The other two Pythagorean identities, 1 + tan²A = sec²A and 1 + cot²A = cosec²ASimplifying products of trigonometric ratios by writing every term in sin and cos firstUsing 30°, 45° and 60° as a numerical check on an algebraic simplification

Common traps

Expanding (1 − cos²A)(1 − sin²A) as 1 − sin²A − cos²A and losing the sin²A cos²A term
Reporting the value of the whole right-hand side instead of the value of m
Multiplying the four factors out in the order printed instead of pairing each reciprocal with its bracket

The expression is built to collapse: each bracket is one Pythagorean substitution away from becoming the reciprocal of the factor standing beside it.

Quant Q.1 of this paper does the same job with (cosec θ − sin θ)(sec θ − cos θ)(tan θ + cot θ), and 23 Sep 2024, 12:30, Quant Q.18 uses cos A + cos²A = 1 to reach sin²A + sin⁴A.

Related PYQs

No directly related past PYQ was found.