Option (a) shows **½{sin(A + B) + sin(A − B)}**. Adding the two sine expansions cancels the cos A sin B terms and leaves 2 sin A cos B, so this expression equals **sin A cos B**, not a product of two sines.

Option (c) shows **½{cos(A + B) − cos(A − B)}** — the same two cosines in the opposite order. That subtraction gives −2 sin A sin B, so the value is **−sin A sin B**, correct in size and wrong in sign.

Option (d) shows **½{sin(A + B) − sin(A − B)}**. Subtracting the sine expansions cancels sin A cos B and leaves 2 cos A sin B, so it equals **cos A sin B**.

Join @UPSCneil — Free Guidance

SSC CGL 2024 Tier-I · 18 September 2024, Shift 3 · Quantitative Aptitude · Question 5 of 98

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.

SinASinB = ______.

Answer

Why

Correct — B. The question is an image reading "SinASinB = ______", and the four options are images of expressions. Start from the two cosine expansions.

cos(A − B) = cos A cos B + sin A sin B
cos(A + B) = cos A cos B − sin A sin B

Subtract the second from the first. The cos A cos B terms cancel and
cos(A − B) − cos(A + B) = 2 sin A sin B

Halve both sides: sin A sin B = ½{cos(A − B) − cos(A + B)}, the expression printed in option (b).

Why the others are wrong

(a)
Option (a) shows ½{sin(A + B) + sin(A − B)}. Adding the two sine expansions cancels the cos A sin B terms and leaves 2 sin A cos B, so this expression equals sin A cos B, not a product of two sines.
(c)
Option (c) shows ½{cos(A + B) − cos(A − B)} — the same two cosines in the opposite order. That subtraction gives −2 sin A sin B, so the value is −sin A sin B, correct in size and wrong in sign.
(d)
Option (d) shows ½{sin(A + B) − sin(A − B)}. Subtracting the sine expansions cancels sin A cos B and leaves 2 cos A sin B, so it equals cos A sin B.

Concept

These are the product-to-sum identities, and all four fall out of the compound-angle formulas in one line each.

Add or subtract cos(A + B) and cos(A − B) and you isolate a product of two cosines or two sines. Add or subtract sin(A + B) and sin(A − B) and you isolate a mixed sine-cosine product.

The useful rule of thumb: a product of two sines or two cosines resolves into cosines, while a mixed product resolves into sines. That alone eliminates two of the four options here before any algebra.

The paper prints the stem as a picture with the spacing "SinASinB", and every option is a picture too, so nothing can be matched by searching the text. Expand one option and compare.

A 20-second numeric check settles it: with A = B = 30°, sin A sin B = 0.25, and ½{cos 0° − cos 60°} = ½(1 − 0.5) = 0.25.

Key facts

2 sin A sin B = cos(A − B) − cos(A + B).
2 cos A cos B = cos(A + B) + cos(A − B).
2 sin A cos B = sin(A + B) + sin(A − B).
2 cos A sin B = sin(A + B) − sin(A − B).

Study next

Product-to-sum and sum-to-product conversionsDeriving the four identities from the compound-angle formulas instead of memorising themSetting A = B to recover the double-angle results as a check

Common traps

Reversing the order to cos(A + B) − cos(A − B), which yields the negative of the right value.
Expecting a sine pair on the right when the product is sine times sine.
Matching the option pictures by shape rather than expanding one of them.

SSC asks identity recall in a single line and sets the options as images, so the work has to be done on paper. A numeric cousin at 17 Sep 2024, 09:00, Quant Q.22 gives tan A = 1 in a right-angled triangle and asks for 4 sin A cos A, again with picture options.

Related PYQs

No directly related past PYQ was found.