Let $\frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)}$, where $A, B, C$ are angles of a triangle $A B C$. If the lengths of the sides opposite these angles are $a, b, c$ respectively, then: [JEE Main 2021, 27 Aug (Shift 1)]

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Correct answer: d) $b^2, c^2, a^2$ are in $A . P$. $ \frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)} $ $\sin A \sin C \cos B-\sin A \cos C \sin B=\sin B \sin A \cos C - \sin B \cos A \sin C $ $ 2 \sin A \sin B \cos C=\sin A \sin C \cos B+\sin B \sin C \cos A $ Using sine rule $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k\ 2⋅ak⋅bk⋅(\frac{{a}^{2}+{b}^{2}−{c}^{2}}{2ab})=ak⋅ck⋅(\frac{{c}^{2}+{a}^{2}−{b}^{2}}{2ac})+bk\cdot ck\left(\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\right)\ \Rightarrow 2({a}^{2}+{b}^{2}−{c}^{2})={c}^{2}+{a}^{2}−{b}^{2}+{b}^{2}+{c}^{2}−{a}^{2}\ \Rightarrow 2{c}^{2}={a}^{2}+{b}^{2}$ $\Rightarrow {b}^{2},{c}^{2},{a}^{2}$ are in A.P

Let \(\frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)}\), where \(A, B, C\) are angles of a triangle \(A B C\). If th…

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