SSC CGL 2024 Tier-I · 18 September 2024, Shift 2 · Quantitative Aptitude · Question 8 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 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.

Simplify. sin A⁄(1 + cos A) + cotA, where A is an acute angle.

(a)sinA+cos A
(b)cos A
(c)sin A
(d)cosec A

Answer

Why

Correct — D. Write cot A as cos A⁄sin A and put both terms over sin A (1 + cos A).

sin A⁄(1 + cos A) + cos A⁄sin A
= [sin²A + cos A (1 + cos A)] ⁄ [sin A (1 + cos A)]
= [sin²A + cos²A + cos A] ⁄ [sin A (1 + cos A)]

Apply sin²A + cos²A = 1:
= (1 + cos A) ⁄ [sin A (1 + cos A)]
= 1⁄sin A = cosec A → option (d).

Why the others are wrong

(a)
sinA+cos A — sin A + cos A comes from adding numerators and denominators separately. The two fractions carry different denominators and need a common denominator first.
(b)
cos A — cos A survives nowhere. It leaves in the last step, when the factor (1 + cos A) cancels top and bottom.
(c)
sin A — sin A is the reciprocal of the answer. The working ends at 1⁄sin A, and that is cosec A, not sin A.

Concept

Two standard forms of the same expression are worth holding:

sin A⁄(1 + cos A) = (1 − cos A)⁄sin A

Multiply the left form top and bottom by (1 − cos A), use 1 − cos²A = sin²A, and one factor of sin A cancels.

That opens a two-line route: (1 − cos A)⁄sin A + cos A⁄sin A = (1 − cos A + cos A)⁄sin A = 1⁄sin A. Both forms also equal tan(A⁄2), which carries straight into half-angle work.

The rider 'where A is an acute angle' is not decoration. It keeps sin A non-zero and 1 + cos A non-zero, so both cancellations are legal — at A = 180° the denominator 1 + cos A would vanish.

Key facts

sin²A + cos²A = 1 is what collapses the numerator to 1 + cos A.
sin A⁄(1 + cos A) equals (1 − cos A)⁄sin A, and both equal tan(A⁄2).
cosec A is defined as 1⁄sin A.

Study next

Rationalising a trigonometric denominator by multiplying with its conjugateThe half-angle forms tan(A⁄2) = sin A⁄(1 + cos A) = (1 − cos A)⁄sin ARewriting cot A and tan A in sine and cosine before combining fractions

Common traps

Adding the two fractions without first taking a common denominator
Cancelling sin A against sin²A before the cos A term has been brought in
Stopping at 1⁄sin A when the options are named as ratios

This one arrives entirely as a picture — the word Simplify, the expression and the acute-angle rider are all inside the image, and the stem text on screen is empty.

The same sub-expression is asked on its own at 9 Sep 2024, 16:00, Quant Q.19: which of the following is equal to sin A ⁄ (1 + cos A), for A acute. That is this card's identity with the cot A term stripped away.

A companion collapse-to-one-ratio item runs at 25 Sep 2024, 09:00, Quant Q.1, where (cosec θ − sin θ)(sec θ − cos θ)(tan θ + cot θ) reduces to a single value.

Related PYQs

No directly related past PYQ was found.