Question

With usual notations in $\triangle \text{ABC}$, if $\frac{\sin \text{A}}{\sin \text{C}} = \frac{\sin(\text{A} - \text{B})}{\sin(\text{B} - \text{C})}$, then $a^2, b^2, c^2$ are in

• A. Not in AP
• B. HP
• C. AP
• D. GP

Hint:

Whenever an identity simplifies to a symmetric squared term relation like $\sin^2\text{B} = 2\sin\text{A}\sin\text{C}\cos\text{B}$, it almost always indicates an underlying Arithmetic Progression for the squares of the sides ($2b^2 = a^2 + c^2$). Memorizing this standard identity trap can save massive amounts of expansion algebra!

Answer 1

The Correct Option is C

Step 1: Clear the given ratio.
Cross-multiplying $\dfrac{\sin A}{\sin C} = \dfrac{\sin(A-B)}{\sin(B-C)}$ gives $\sin A\,\sin(B-C) = \sin C\,\sin(A-B)$.
Step 2: Replace $A$ and $C$ cleverly.
Since $A + B + C = \pi$, we have $A = \pi - (B+C)$ so $\sin A = \sin(B+C)$, and similarly $\sin C = \sin(A+B)$.
Step 3: Substitute these.
The equation becomes $\sin(B+C)\sin(B-C) = \sin(A+B)\sin(A-B)$.
Step 4: Use $\sin(P+Q)\sin(P-Q) = \sin^2 P - \sin^2 Q$.
This gives $\sin^2 B - \sin^2 C = \sin^2 A - \sin^2 B$, hence $2\sin^2 B = \sin^2 A + \sin^2 C$.
Step 5: Switch to sides using the Sine Rule.
Since $\sin A \propto a$, $\sin B \propto b$, $\sin C \propto c$, the relation becomes $2b^2 = a^2 + c^2$.
Step 6: Recognise the progression.
$2b^2 = a^2 + c^2$ means $a^2,\ b^2,\ c^2$ are in AP.
\[ \boxed{a^2,\ b^2,\ c^2 \text{ are in AP}} \]