SSC CGL 2024 Tier-I · 24 September 2024, Shift 2 · Quantitative Aptitude · Question 20 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.

The value of (sin A + cosec A)² + (cos A + sec A)² is:

Answer

Why

Correct — A. The stem, printed as an image, asks for the value of (sinA + cosecA)² + (cosA + secA)².

Expand both squares:
(sinA + cosecA)² = sin²A + 2·sinA·cosecA + cosec²A
(cosA + secA)² = cos²A + 2·cosA·secA + sec²A

sinA·cosecA = 1 and cosA·secA = 1, so each middle term is 2 — 4 between them.

Now the rest:
sin²A + cos²A = 1
cosec²A = 1 + cot²A and sec²A = 1 + tan²A

Total = 1 + 4 + 1 + 1 + tan²A + cot²A = 7 + tan²A + cot²A, which is option (a).

Why the others are wrong

(b)
Option (b) shows 1 + tan²A + cot²A. Its constant keeps only sin²A + cos²A = 1 and throws away both middle terms and the two 1s hidden inside cosec²A and sec²A — six of the seven.
(c)
Option (c) shows 3 + tan²A + cot²A. It collects sin²A + cos²A = 1 and the two 1s from cosec²A = 1 + cot²A and sec²A = 1 + tan²A, but leaves out the two middle terms, worth 2 each.
(d)
Option (d) shows 5 + tan²A + cot²A, which is 1 + 4 — the Pythagorean 1 plus both middle terms — with cosec²A and sec²A turned straight into cot²A and tan²A, losing the +1 each carries.

Concept

Two families of identity do all the work.

Reciprocal: sinA·cosecA = 1 and cosA·secA = 1. That fixes every middle term at 2 without knowing anything about A.

Pythagorean: sin²A + cos²A = 1, cosec²A = 1 + cot²A, sec²A = 1 + tan²A. The last two are where the constant grows, each adding a 1 on top of its squared ratio.

Add the pieces: 1 from the first identity, 2 + 2 from the reciprocals, 1 + 1 from the last two. The constant is 7, and tan²A + cot²A is the part no identity removes.

All four option images carry the same tail, tan²A + cot²A, so the item is only ever about the constant in front. If the expansion goes wrong, substitute A = 45°: the expression becomes (3⁄√2)² + (3⁄√2)² = 9, and only option (a) returns 9.

Key facts

sinA · cosecA = 1 and cosA · secA = 1, so a middle term 2·sinA·cosecA is simply 2.
cosec²A = 1 + cot²A and sec²A = 1 + tan²A.
(sinA + cosecA)² + (cosA + secA)² = 7 + tan²A + cot²A wherever all four ratios are defined.
The constant 7 is 1 from sin²A + cos²A, 4 from the two middle terms, and 1 + 1 from the two Pythagorean forms.

Study next

The three Pythagorean identities and every rearrangement of themReciprocal pairs: sine with cosec, cosine with sec, tan with cotExpanding (a + b)² where the product ab is a known constant

Common traps

Dropping the two middle terms, which costs 4 from the constant.
Converting cosec²A to cot²A without carrying the extra 1, and sec²A to tan²A without carrying its own.
Expanding the first bracket carefully and assuming the second behaves the same way without checking.

Squaring a compound trigonometric expression and reducing it with the Pythagorean forms is a standing SSC shape. Quant Q.19 of this paper wants the same squaring on tanA + cotA, and Quant Q.25 reaches sine and cosine from a single given ratio instead.

Related PYQs

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