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Question
If \[2 \tan \alpha = 3 \tan \beta, \text{ then } \tan \left( \alpha - \beta \right) =\]
Options
\[\frac{\sin 2 \beta}{5 - \cos 2 \beta}\]
- \[\frac{\cos 2 \beta}{5 - \cos 2 \beta}\]
\[\frac{\sin 2 \beta}{5 + \cos 2 \beta}\]
none of these
Solution
\[\frac{\sin 2 \beta}{5 - \cos 2 \beta}\]
\[\text{ Given } : \]
\[2\tan\alpha = 3\tan\beta\]
\[\text{ Now } , \]
\[\tan\left( \alpha - \beta \right) = \frac{tan\alpha - tan\beta}{1 + \text{ tan } \alpha \text{ tan } \beta}\]
\[ = \frac{\frac{3}{2}tan\beta - tan\beta}{1 + \left( \frac{3}{2}tan\beta \right)tan\beta}\]
\[ = \frac{3tan\beta - 2tan\beta}{2 + 3 \tan^2 \beta}\]
\[ = \frac{tan\beta}{2 + 3 \tan^2 \beta}\]
\[ = \frac{\frac{sin\beta}{cos\beta}}{2 + 3\frac{\sin^2 \beta}{\cos^2 \beta}}\]
\[ = \frac{sin\beta cos\beta}{2 \cos^2 \beta + 3 \sin^2 \beta}\]
\[ = \frac{sin\beta cos\beta}{2 + \sin^2 \beta}\]
\[ = \frac{2sin\beta cos\beta}{4 + 2 \sin^2 \beta}\]
\[ = \frac{\sin2\beta}{4 + 1 - \cos2\beta}\]
\[ \therefore \tan\left( \alpha - \beta \right) = \frac{\sin2\beta}{5 - \cos2\beta}\]
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