With usual notations prove that (A-B) (A+B) =a^2-b^2 c^2 . - Vidyadip

Question

With usual notations prove that $\frac{\sin (A-B)}{\sin (A+B)}=\frac{a^2-b^2}{c^2}$.

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Answer

By the sine rule,
$\begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \\ & \therefore a=k \sin A, b=k \sin B, c=k \sin C\end{aligned}$
$\mathrm{RHS}=\frac{a^2-b^2}{c^2}=\frac{k^2 \sin ^2 \mathrm{~A}-k^2 \sin ^2 \mathrm{~B}}{k^2 \sin ^2 \mathrm{C}}$
$=\frac{\sin ^2 A-\sin ^2 B}{\sin ^2 C}$
$=\frac{(\sin A+\sin B)(\sin A-\sin B)}{[\sin \{\pi-(A+B)\}]^2}$
$\ldots[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi]$
$=\frac{2 \sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right) \times 2 \cos \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right)}{\sin ^2(A+B)}$
$=\frac{2 \sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A+B}{2}\right) \times 2 \sin \left(\frac{A-B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)}{\sin ^2(A+B)}$
$=\frac{\sin (A+B) \cdot \sin (A-B)}{\sin ^2(A+B)}$
$=\frac{\sin (A-B)}{\sin (A+B)}=$ LHS.

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