SSC CGL 2024 Tier-I · 12 September 2024, Shift 2 · Quantitative Aptitude · Question 22 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 6 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 cot⁴φ − cot²φ = 1, then the value of cos⁴φ + cos²φ is:

Answer

Why

Correct — A. The stem is an image: if cot⁴φ − cot²φ = 1, find cos⁴φ + cos²φ. Move the cot² term across and the identity 1 + cot²φ = cosec²φ appears.

cot⁴φ − cot²φ = 1
cot⁴φ = 1 + cot²φ = cosec²φ

Write both sides in sine and cosine:
cos⁴φ ⁄ sin⁴φ = 1 ⁄ sin²φ
Multiply through by sin⁴φ → cos⁴φ = sin²φ

Substitute sin²φ = 1 − cos²φ:
cos⁴φ = 1 − cos²φ
cos⁴φ + cos²φ = 1 → option (a), the figure showing 1.

Why the others are wrong

(b)
Option (b) shows 3⁄2. Once cos⁴φ = sin²φ, the quantity asked for is sin²φ + cos²φ, and that Pythagorean identity is exactly 1 — it can never exceed it.
(c)
Option (c) shows 1⁄√3, which is cot 60°. Substituting a convenient angle fails here: at 60°, cot⁴φ − cot²φ = 1⁄9 − 1⁄3, which is not 1, so 60° does not satisfy the condition.
(d)
Option (d) shows 2, which is what you get by treating cos⁴φ and cos²φ as 1 each. Only their sum is pinned to 1; taken separately both are smaller than 1.

Concept

Three Pythagorean identities do the work in an identity item like this one: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ.

A condition such as cot⁴φ − cot²φ = 1 is written so that it rearranges into one of them, here cot⁴φ = 1 + cot²φ = cosec²φ. Converting to sine and cosine then collapses it to cos⁴φ = sin²φ.

The expression you are asked for is chosen so that this one substitution turns it back into sin²φ + cos²φ, which is 1 for every angle that satisfies the condition.

Both the stem and the four options are images on the response sheet rather than text. Option (a) shows 1, option (b) 3⁄2, option (c) 1⁄√3 and option (d) 2.

Key facts

1 + cot²θ = cosec²θ is the identity that turns the given condition into cot⁴φ = cosec²φ.
In sine and cosine that reads cos⁴φ ⁄ sin⁴φ = 1 ⁄ sin²φ, so cos⁴φ = sin²φ.
Substituting sin²φ = 1 − cos²φ gives cos⁴φ + cos²φ = 1.
The mirrored statement is also true: if cos A + cos²A = 1, then sin²A + sin⁴A = 1.

Study next

The mirrored problem, cos A + cos²A = 1 leading to sin²A + sin⁴ARearranging a given condition into 1 + tan²θ = sec²θ before substitutingRewriting tan, cot, sec and cosec in sine and cosine to collapse a fraction

Common traps

Plugging in a standard angle: 30°, 45° and 60° all fail cot⁴φ − cot²φ = 1.
Reading cot⁴φ as cot 4φ rather than as the fourth power of cot φ.
Stopping at cos⁴φ = sin²φ and reporting sin²φ instead of the sum asked for.

SSC gives a condition that hides one Pythagorean identity, then asks for a second expression that the condition collapses. The mirrored version, if cos A + cos²A = 1 find sin²A + sin⁴A, is asked on 23 Sep 2024, 12:30, Quant Q.18.

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