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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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