SSC CGL 2024 Tier-I · 9 September 2024, Shift 3 · Quantitative Aptitude · Question 19 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 A is an acute angle, which of the following is equal to sin A ⁄ (1 + cos A)?

Answer

Why

Correct — D. The stem is printed as an image: for an acute angle A, which option equals sin A ⁄ (1 + cos A)?

Multiply numerator and denominator by the conjugate (1 − cos A):

= sin A(1 − cos A) ⁄ [(1 + cos A)(1 − cos A)]
denominator = 1 − cos²A = sin²A
= sin A(1 − cos A) ⁄ sin²A
cancel one sin A → (1 − cos A) ⁄ sin A → option (d)

Check at A = 60°: sin 60°⁄(1 + cos 60°) = 0.866⁄1.5 = 0.577, and (1 − 0.5)⁄0.866 = 0.577.

Why the others are wrong

(a)
Option (a) shows (1 + sin A) ⁄ cos A, which is sec A + tan A. At A = 60° it reads 3.73 against the stem's 0.577 — the wrong pair of functions altogether.
(b)
(1 − sin A) ⁄ cos A is what you get by rationalising cos A⁄(1 + sin A), a different fraction. At A = 60° it gives 0.268, well below the stem's 0.577.
(c)
Option (c), (1 + cos A) ⁄ sin A, is the reciprocal of the given fraction — cot(A⁄2) where the stem is tan(A⁄2). At A = 60° it reads 1.732.

Concept

Every rationalise-the-trig-fraction item runs on sin²A + cos²A = 1, used in its difference-of-squares form.

Multiplying (1 + cos A) by its conjugate (1 − cos A) leaves 1 − cos²A, which is sin²A — a clean single-function denominator. The mirror move works on (1 ± sin A), leaving cos²A.

Two half-angle results are worth carrying outright: sin A⁄(1 + cos A) = (1 − cos A)⁄sin A = tan(A⁄2), and (1 + cos A)⁄sin A = cot(A⁄2).

Recognising the pair saves the algebra entirely.

The stem specifies an acute angle so nothing is zero or negative: sin A > 0 keeps the cancellation legal and fixes the sign of the result, and it also makes a numerical check at A = 60° safe.

Key facts

(1 + cos A)(1 − cos A) = 1 − cos²A = sin²A.
sin A⁄(1 + cos A) and (1 − cos A)⁄sin A are the same expression, both equal to tan(A⁄2).
Its reciprocal, (1 + cos A)⁄sin A, equals cot(A⁄2).
sec A + tan A can be written as (1 + sin A)⁄cos A, which is a different family.

Study next

Conjugate multiplication on (1 ± sin θ) and (1 ± cos θ) denominatorsHalf-angle forms of tan and cot written in sin A and cos AVerifying a trig identity by substituting a convenient acute angle

Common traps

Multiplying by (1 + cos A) rather than the conjugate, which leaves (1 + cos A)² and goes nowhere.
Cancelling both sin A factors and losing the denominator entirely.
Answering the reciprocal after rationalising the fraction the wrong way up.

SSC prints the fraction as an image and surrounds it with four look-alike fractions, so substituting A = 60° is the fastest safe check. This shift also sets an identity item of the same family at Quant Q.22, where sec α and tan α have to be rewritten in sin α.

Related PYQs

No directly related past PYQ was found.