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Question
If \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\] is equal to
Options
\[2 \tan \beta\]
\[3 \tan \beta\]
\[4 \tan \beta\]
\[6 \tan \beta\]
Solution
\[4 \tan \beta\]
\[ \text{ We have } , \]
\[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right)\]
\[ \Rightarrow \frac{5}{3} = \frac{\sin \left( \alpha + 2 \beta \right)}{\sin \alpha}\]
\[ \Rightarrow \frac{5 - 3}{5 + 3} = \frac{\sin \left( \alpha + 2 \beta \right) - \sin \alpha}{\sin \left( \alpha + 2 \beta \right) + \sin \alpha} \left( \text{ Using componendo and dividendo } \right)\]
\[ \Rightarrow \frac{2}{8} = \frac{\sin \left( \alpha + 2 \beta \right) - \sin \alpha}{\sin \left( \alpha + 2 \beta \right) + \sin \alpha}\]
\[ \Rightarrow \frac{1}{4} = \frac{2\cos\frac{\alpha + 2 \beta + \alpha}{2}\sin\frac{\alpha + 2 \beta - \alpha}{2}}{2\sin\frac{\alpha + 2 \beta + \alpha}{2}\cos\frac{\alpha + 2 \beta - \alpha}{2}}\]
\[ \Rightarrow \frac{1}{4} = \frac{\cos\left( \alpha + \beta \right) \sin \beta}{\sin\left( \alpha + \beta \right) \cos \beta}\]
\[ \Rightarrow \frac{1}{4} = \cot \left( \alpha + \beta \right) \tan \beta\]
\[ \Rightarrow \frac{1}{4} = \frac{1}{\tan \left( \alpha + \beta \right)}\tan \beta\]
\[ \therefore \tan \left( \alpha + \beta \right) = 4 \tan \beta\]
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