Question
In ∆ABC prove the following : (i) a sin A – b sin B = c sin (A – B)
$\begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \\ & \therefore a=k \sin A, b=k \sin B, c=k \sin C_c\end{aligned}$
LHS $=a \sin A-b \sin B$
$\begin{aligned} & =k \sin A \cdot \sin A-k \sin B \cdot \sin B \\ & =k\left(\sin ^2 A-\sin ^2 B\right)\end{aligned}$
$=k(\sin A+\sin B)(\sin A-\sin B)$
$\begin{aligned} &=k \times 2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \times \\ & 2 \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\end{aligned}$
$\begin{aligned}=k \times 2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \times \\ 2 \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\end{aligned}$
$\begin{aligned} & =k \times \sin (A+B) \times \sin (A-B) \\ & =k \sin (\pi-C) \cdot \sin (A-B) \ldots[\because A+B+C=\pi] \\ & =k \sin C \cdot \sin (A-B) \\ & =c \sin (A-B)=\text { RHS. }\end{aligned}$
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