Let $\frac{\sin \mathrm{A}}{\sin \mathrm{B}}=\frac{\sin (\mathrm{A}-\mathrm{C})}{\sin (\mathrm{C}-\mathrm{B})}$, where $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are angles of a triangle $\mathrm{ABC}$. If the lengths of the sides opposite these angles are $a, b, c$ respectively, then :
Correct Option: , 2
$\frac{\sin \mathrm{A}}{\sin \mathrm{B}}=\frac{\sin (\mathrm{A}-\mathrm{C})}{\sin (\mathrm{C}-\mathrm{B})}$
As A,B,C are angles of triangle
$A+B+C=\pi$
$A=\pi-(B+C)$
So, $\sin A=\sin (B+C) \ldots$ (1)
Similarly $\sin B=\sin (A+C) \ldots(2)$
From (1) and (2)
$\frac{\sin (\mathrm{B}+\mathrm{C})}{\sin (\mathrm{A}+\mathrm{C})}=\frac{\sin (\mathrm{A}-\mathrm{C})}{\sin (\mathrm{C}-\mathrm{B})}$
$\sin (C+B) \cdot \sin (C-B)=\sin (A-C) \sin (A+C)$
$\sin ^{2} \mathrm{C}-\sin ^{2} \mathrm{~B}=\sin ^{2} \mathrm{~A}-\sin ^{2} \mathrm{C}$
$\left\{\because \sin (x+y) \sin (x-y)=\sin ^{2} x-\sin ^{2} y\right\}$
$2 \sin ^{2} \mathrm{C}=\sin ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~B}$
By sine rule
$2 c^{2}=a^{2}+b^{2}$
$\Rightarrow \mathrm{b}^{2}, \mathrm{c}^{2}$ and $\mathrm{a}^{2}$ are in A.P.
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