Question

In triangle ABC, prove the following:

$\frac{a^2-c^2}{b^2}=\frac{\sin \left(A-C\right)}{\sin \left(A+C\right)}$

Answer 1

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$
Then,
Consider the LHS of the equation $\frac{a^2-c^2}{b^2}=\frac{\sin \left(A-C\right)}{\sin \left(A+C\right)}$.
$$\begin{gathered} \mathrm{LHS}=\frac{{\left[k\sin \left(\mathrm{A}\right)\right]}^2-{\left[\mathrm{k}\sin \left(\mathrm{C}\right)\right]}^2}{{\left(\mathrm{k}\sin \left(\mathrm{B}\right)\right)}^2} \\ =\frac{k^2\left({\sin }^2\left(A\right)-{\sin }^2\left(C\right)\right)}{k^2{\sin }^2\left(B\right)} \\ =\frac{\sin \left(A+C\right)\sin \left(A-C\right)}{{\sin }^{2}B}\left[\because {\sin }^{2}A-{\sin }^{2}C=\sin \left(A+C\right)\sin \left(A-C\right)\right] \\ =\frac{\sin \left(A+C\right)\sin \left(A-C\right)}{Si{n}^2\left(\mathrm{\pi }-\left(\mathrm{A}+\mathrm{C}\right)\right)}\left[\because \left(A+B+C\right)=\mathrm{\pi }\right] \\ =\frac{\sin \left(A+C\right)\sin \left(A-C\right)}{{\sin }^2\left(A+C\right)} \\ =\frac{\sin \left(A-C\right)}{\sin \left(A+C\right)}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$