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In Triangle Abc, Prove the Following: a 2 − C 2 B 2 = Sin ( a − C ) Sin ( a + C ) - Mathematics

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Question

In triangle ABC, prove the following: 

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

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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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Sine and Cosine Formulae and Their Applications
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Chapter 10: Sine and cosine formulae and their applications - Exercise 10.1 [Page 13]

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RD Sharma Mathematics [English] Class 11
Chapter 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 10 | Page 13

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