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Question
If \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
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Solution
Given: \[2 \tan \alpha = 3 \tan \beta\]
\[ = \frac{\frac{3}{2} \times tan\beta - tan\beta}{1 + \frac{3}{2} \tan^2 \beta} \left( \because 2tan\alpha = 3tan\beta \right)\]
\[ = \frac{\frac{1}{2} \times tan\beta}{1 + \frac{3}{2} \tan^2 \beta} = \frac{tan\beta}{2 + 3 \tan^2 \beta}\]
\[ = \frac{\frac{sin\beta}{cos\beta}}{2 + 3\frac{\sin^2 \beta}{\cos^2 \beta}} = \frac{\frac{sin\beta}{cos\beta} \times \cos^2 \beta}{2 \cos^2 \beta + 3 \sin^2 \beta}\]
\[ = \frac{1}{2}\frac{2sin\beta \times cos\beta}{2\left( \cos^2 \beta + \sin^2 \beta \right) + \sin^2 \beta}\]
\[ = \frac{1}{2}\frac{\sin2\beta}{\left( 2 + \sin^2 \beta \right)} = \frac{\sin2\beta}{4 + 2 \sin^2 \beta}\]
\[ = \frac{\sin2\beta}{4 + 2\left( 1 - \cos^2 \beta \right)} = \frac{\sin2\beta}{6 - 2 \cos^2 \beta}\]
\[ = \frac{\sin2\beta}{6 - \left( 1 + \cos2\beta \right)} \left( \because 1 + \cos2\beta = 2 \cos^2 \beta \right)\]
\[ = \frac{\sin2\beta}{5 - \cos2\beta} = RHS\]
\[\text{ Hence proved } .\]
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