SSC CGL 2024 Tier-I · 11 September 2024, Shift 2 · Quantitative Aptitude · Question 8 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.

Find the value of: [(4⁄3) tan²60° + 3 cos²30° − 2 sec²30° − (3⁄4) cot²60°] ÷ [sin 60° cos 30° − cos 60° sin 30°]

Answer

Why

Correct — C. The stem and the options are images. The expression is
[ (4⁄3)tan²60° + 3cos²30° − 2sec²30° − (3⁄4)cot²60° ] ÷ [ sin60°cos30° − cos60°sin30° ]

Evaluate the four bracket terms from the standard table:

(4⁄3)tan²60° = (4⁄3)(3) = 4
3cos²30° = 3(3⁄4) = 9⁄4
2sec²30° = 2(4⁄3) = 8⁄3
(3⁄4)cot²60° = (3⁄4)(1⁄3) = 1⁄4

Numerator = 4 + 9⁄4 − 8⁄3 − 1⁄4
The two quarter-terms combine to 2, so this is 6 − 8⁄3 = 10⁄3

Denominator: sin A cos B − cos A sin B = sin(A − B), so it is sin(60° − 30°) = sin30° = 1⁄2

(10⁄3) ÷ (1⁄2) = (10⁄3) × 2 = 20⁄3 = 6 2⁄3 → option (c).

Why the others are wrong

(a)
3 1⁄2 is 7⁄2, which would need a numerator of 7⁄4. The bracket gives 4 + 9⁄4 − 8⁄3 − 1⁄4, where the quarter-terms combine to 2 and leave 6 − 8⁄3 = 10⁄3.
(b)
4 1⁄3 is 13⁄3, which would need a numerator of 13⁄6. Nothing in the bracket produces sixths — the denominators in play are thirds and quarters, landing on 10⁄3.
(d)
6 3⁄2 is a mixed number with an improper fractional part, worth 7.5. The division gives 20⁄3, and 20⁄3 as a mixed number is 6 2⁄3 — the digits transposed.

Concept

Two separate skills are stacked here, which is why the question looks long.

The bracket is pure table work: every term is a standard angle, and the only care needed is that sec and cot are reciprocals of cos and tan, so their squares invert.

The divisor is an identity in disguise. Whenever you see sin A cos B − cos A sin B, read it as sin(A − B) rather than multiplying four values.

Then finish carefully: dividing by 1⁄2 means multiplying by 2.

The identity is a shortcut, not a necessity. Multiplied out, the divisor is (√3⁄2)(√3⁄2) − (1⁄2)(1⁄2) = 3⁄4 − 1⁄4 = 1⁄2, exactly what sin30° gives.

If you cannot recall the compound-angle form under pressure, four standard values get you there in the same place.

Key facts

tan 60° = √3, so tan²60° = 3.
sec 30° = 2 ⁄ √3, so sec²30° = 4⁄3.
cot 60° = 1 ⁄ √3, so cot²60° = 1⁄3.
sin A cos B − cos A sin B = sin(A − B), so sin60°cos30° − cos60°sin30° = sin30° = 1⁄2.

Study next

The compound-angle identities for sin, cos and tanThe Pythagorean identities sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θThe full table of all six ratios at 0°, 30°, 45°, 60° and 90°

Common traps

Dividing by 1⁄2 instead of multiplying by 2
Confusing sec²30° = 4⁄3 with cosec²30° = 4
Reading the familiar pattern as sin(A + B) and getting sin90° = 1

SSC builds one long bracket out of standard angles and hides a compound-angle identity in the divisor, so the length is intimidation rather than difficulty. Trigonometry is asked again in this shift at Quant Q.9, where sin A = 1⁄2 in a right-angled triangle fixes cos C.

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