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In ∆Abc, Prove the Following: ( C 2 − a 2 + B 2 ) Tan a = ( a 2 − B 2 + C 2 ) Tan B = ( B 2 − C 2 + a 2 ) Tan C

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Question

In ∆ABC, prove the following

\[\left( c^2 - a^2 + b^2 \right) \tan A = \left( a^2 - b^2 + c^2 \right) \tan B = \left( b^2 - c^2 + a^2 \right) \tan C\] 

 

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Solution

\[\text{ We know that }\]

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k\]

\[\text{ So }, \]

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

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

\[ = 2bc\sin A\]

\[ = 2kabc . . . \left( 1 \right)\]

\[\left( a^2 - b^2 + c^2 \right)\tan B = \left( a^2 - b^2 + c^2 \right)\frac{\sin B}{\cos B}\]

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

\[ = 2ac\sin B\]

\[ = 2kabc . . . \left( 2 \right)\]

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

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

\[ = 2ab\sin C\]

\[ = 2kabc . . . . \left( 3 \right)\]

From (1), (2) and (3), we get: 

\[\left( c^2 - a^2 + b^2 \right) \tan A = \left( a^2 - b^2 + c^2 \right) \tan B = \left( b^2 - c^2 + a^2 \right) \tan C\]

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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 10.2 [Page 25]

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RD Sharma Mathematics [English] Class 11
Chapter 10 Sine and cosine formulae and their applications
Exercise 10.2 | Q 8 | Page 25

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