Question

In any triangle $ABC$ if $\frac{\sin A}{\sin C}=\frac{\sin A-B}{\sin B-C}$ prove that $a^2,b^2$ and $c^2$ are in A.P.

Answer 1

In $△ABC,A+B+C=\pi$

$\Rightarrow A=\pi -(B+C)$ and $C=\pi -(A+B)$

Now, $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$

$\Rightarrow \frac{\sin (\pi -(B+C))}{\sin (\pi -(A+B))}=\frac{\sin (A-B)}{\sin (B-C)}$

$\Rightarrow \frac{\sin (B+C)}{\sin (A+B)}=\frac{\sin (A-B)}{\sin (B-C)}$

$\Rightarrow \sin (B+C)\sin (B+C)=\sin (A-B)\sin (A+B)$

$\Rightarrow {\sin }^{2}B-{\sin }^{2}C={\sin }^{2}A-{\sin }^{2}B$

By sine rule,

$\Rightarrow b^2{k}^2-c^2{k}^2=a^2{k}^2-b^2{k}^2$

$\Rightarrow b^2-c^2=a^2-b^2$

$\therefore a^2,b^2,c^2$ are in A.P