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Question
In triangle ABC, prove the following:
\[\frac{a^2 - c^2}{b^2} = \frac{\sin \left( A - C \right)}{\sin \left( A + C \right)}\]
Solution
Let
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]
Then,
Consider the LHS of the equation\[\frac{a^2 - c^2}{b^2} = \frac{\sin \left( A - C \right)}{\sin \left( A + C \right)}\]
\[LHS = \frac{\left[ k\sin\left( A \right) \right]^2 - \left[ k\sin\left( C \right) \right]^2}{\left( k\sin\left( B \right) \right)^2}\]
\[ = \frac{k^2 \left( \sin^2 \left( A \right) - \sin^2 \left( C \right) \right)}{k^2 \sin^2 \left( B \right)}\]
\[ = \frac{\sin\left( A + C \right)\sin\left( A - C \right)}{\sin^2 B} \left[ \because \sin^2 A - \sin^2 C = \sin\left( A + C \right)\sin\left( A - C \right) \right]\]
\[ = \frac{\sin\left( A + C \right)\sin\left( A - C \right)}{Si n^2 \left( \pi - \left( A + C \right) \right)} \left[ \because \left( A + B + C \right) = \pi \right]\]
\[ = \frac{\sin\left( A + C \right)\sin\left( A - C \right)}{\sin^2 \left( A + C \right)}\]
\[ = \frac{\sin\left( A - C \right)}{\sin\left( A + C \right)} = RHS\]
\[ \]
\[\text{ Hence proved } .\]
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