SSC CGL 2024 Tier-I · 19 September 2024, Shift 3 · Quantitative Aptitude · Question 18 of 99

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, then √[(1 − cosA) ⁄ (1 + cosA)] + √[(1 + cosA) ⁄ (1 − cosA)] is equal to:

Answer

Why

Correct — A. The two surds are reciprocals of each other, so put them over one common denominator instead of evaluating either.

Let S = √((1 − cosA)⁄(1 + cosA)) + √((1 + cosA)⁄(1 − cosA))

Common denominator = √((1 + cosA)(1 − cosA)) = √(1 − cos²A) = sinA
Numerator = (1 − cosA) + (1 + cosA) = 2

S = 2 ⁄ sinA = 2 cosecA → option (a), the picture reading 2 cosecA

Why the others are wrong

(b)
2 secA is 2 ⁄ cosA. It comes from taking the common denominator as cosA, but (1 + cosA)(1 − cosA) is 1 − cos²A = sin²A, so sinA is what sits underneath.
(c)
2 cosA multiplies by cosA where the working divides by sinA. It is also below 2 for every acute A, and a positive number plus its reciprocal is never less than 2.
(d)
2 sinA puts sinA in the numerator, inverting the last step of S = 2 ⁄ sinA. Like 2 cosA it stays under 2, which the sum of two reciprocal surds cannot do.

Concept

Anything of the form √((1 − cosA)⁄(1 + cosA)) is cleared by the difference of squares: multiply the two conjugates and (1 + cosA)(1 − cosA) = sin²A.

Because the second surd is the first one upside down, you never have to simplify either separately. Adding p and 1⁄p over a common denominator collapses the numerator to (1 − cosA) + (1 + cosA) = 2.

A second route reaches the same place: each surd is a half-angle form, tan(A⁄2) and cot(A⁄2), and tanθ + cotθ = 1 ⁄ (sinθ cosθ) = 2 ⁄ sinA when θ = A⁄2.

The wording A is an acute angle is doing real work: it makes sinA positive, so √(sin²A) is sinA and not −sinA, and it also rules out A = 0°, where 1 − cosA would be 0 and the second surd undefined.

Key facts

(1 + cosA)(1 − cosA) = 1 − cos²A = sin²A.
cosecA means 1 ⁄ sinA, so 2 ⁄ sinA is written 2 cosecA.
For an acute A, sinA > 0, so √(sin²A) = sinA with no sign ambiguity.
For any positive p, p + 1⁄p is at least 2, which is why the answer cannot be 2 sinA or 2 cosA.

Study next

Rationalising √((1 − cosA)⁄(1 + cosA)) by multiplying numerator and denominator by (1 − cosA)The half-angle identity √((1 − cosA)⁄(1 + cosA)) = tan(A⁄2)Reciprocal pairs sinA·cosecA = 1 and cosA·secA = 1 used to name a simplified result

Common traps

Adding the two surds as if √p + √q equalled √(p + q).
Hesitating over ±sinA when the word acute already fixes the sign as positive.
Rationalising only one surd and losing the symmetry that collapses the numerator to 2.

One expression to simplify, four one-term options: the same build is set at 25 Sep 2024, 09:00, Quant Q.1, which asks for the value of (cosec θ − sin θ)(sec θ − cos θ)(tan θ + cot θ).

A related manipulation, secθ + tanθ = x solved for sinθ, is asked at 25 Sep 2024, 16:00, Quant Q.14.

Related PYQs

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