SSC CGL 2024 Tier-I · 26 September 2024, Shift 2 · Quantitative Aptitude · Question 25 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.

In a ΔABC, right angled at B if tan C = √3, then find (sin²C + cos²C) ⁄ (1 + cot²C).

Answer

Why

Correct — B. The stem and all four options are printed as images: in ΔABC right-angled at B, tan C = √3, and the expression wanted is (sin²C + cos²C)⁄(1 + cot²C).

Numerator: sin²C + cos²C = 1
Denominator: 1 + cot²C = cosec²C

So the whole fraction is 1⁄cosec²C = sin²C.

tan C = √3 → cot C = 1⁄√3, so 1 + cot²C = 1 + 1⁄3 = 4⁄3
Value = 1 ÷ 4⁄3 = 3⁄4 → option (b)

Why the others are wrong

(a)
16⁄3 is greater than 1, which this expression can never be — it reduces to sin²C, and a squared sine cannot exceed 1.
(c)
3⁄16 is sin²C × cos²C, that is 3⁄4 × 1⁄4. It comes from multiplying the two squares in the numerator instead of using sin²C + cos²C = 1.
(d)
4⁄15 carries a denominator of 15, and nothing here produces one: the values in play are sin²C = 3⁄4, cos²C = 1⁄4 and cot²C = 1⁄3.

Concept

Two Pythagorean identities do all the work, and neither of them needs the angle.

sin²θ + cos²θ = 1 collapses the numerator to 1, and 1 + cot²θ = cosec²θ collapses the denominator. What survives is 1⁄cosec²C, which is sin²C.

Only now does tan C = √3 matter, and only through cot C = 1⁄√3. Recognising C = 60° and using sin 60° = √3⁄2 reaches the same 3⁄4 one step later.

The right angle at B is not needed for the identities — they hold at every angle — but it does guarantee C is acute, so tan C = √3 pins C to 60°.

Key facts

sin²θ + cos²θ = 1 at every angle θ.
1 + cot²θ = cosec²θ, so the denominator here is cosec²C.
The whole expression therefore equals sin²C, whatever C turns out to be.
With C acute, tan C = √3 means C = 60°, and sin²60° = 3⁄4.

Study next

The three Pythagorean identities and which one clears which fractionStandard values at 0°, 30°, 45°, 60° and 90°Expressions that reduce to a constant whatever angle is supplied

Common traps

Reaching for 1 + tan²C = sec²C, which makes the denominator 4 and the answer 1⁄4.
Computing sin C and cos C separately when the numerator is already 1.
Reading cot C as √3 rather than as its reciprocal 1⁄√3.

SSC buries a one-line identity inside a fraction and supplies a ratio you may not even need.

The same shape — a right triangle, one ratio given, an expression to evaluate — runs at 17 Sep 2024, 09:00, Quant Q.22, where tan A = 1 leads into 4 sin A cos A.

Related PYQs

No directly related past PYQ was found.