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

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Question

In triangle ABC, prove the following: 

\[b \sin B - c \sin C = a \sin \left( B - C \right)\]

 

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Solution

Let 

\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\] 

Consider the LHS of he equation

\[b \sin B - c \sin C = a \sin \left( B - C \right)\]

LHS \[= k\sin B\sin B - k\sin C\sin C\]

\[= k\left( \sin^2 B - \sin^2 C \right)\]
\[ = k\left[ \sin\left( B + C \right)\sin\left( B - C \right) \right] \left[ \because \sin^2 B - \sin^2 C = \sin\left( B + C \right)\sin\left( B - C \right) \right]\]
\[ = k\left[ \sin\left( \pi - A \right)\sin\left( B - C \right) \right] \left[ \because A + B + C = \pi \right]\]
\[ = k\sin A\sin\left( B - C \right) \left[ \because a = k\sin A \right]\]
\[ = a\sin\left( B - C \right) = RHS\]
\[\text{ Hence proved }.\]

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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 11 | Page 13

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