If \(5\sin\alpha = 3\sin(\alpha + 2\beta)\neq 0\), Find \( \tan(\alpha+\beta) \)
Question
If
\[ 5\sin\alpha=3\sin(\alpha+2\beta)\neq0, \]
then
\[ \tan(\alpha+\beta) \]
is equal to
(a) \(2\tan\beta\)
(b) \(3\tan\beta\)
(c) \(4\tan\beta\)
(d) \(6\tan\beta\)
Solution
Given,
\[ 5\sin\alpha=3\sin(\alpha+2\beta) \]
Using
\[ \sin(\alpha+2\beta) = \sin\alpha\cos2\beta + \cos\alpha\sin2\beta \]
Substituting,
\[ 5\sin\alpha = 3(\sin\alpha\cos2\beta+\cos\alpha\sin2\beta) \]
\[ (5-3\cos2\beta)\sin\alpha = 3\sin2\beta\cos\alpha \]
Dividing by \(\cos\alpha\),
\[ \tan\alpha = \frac{3\sin2\beta}{5-3\cos2\beta} \]
Let
\[ t=\tan\beta \]
Using
\[ \sin2\beta=\frac{2t}{1+t^2}, \qquad \cos2\beta=\frac{1-t^2}{1+t^2} \]
\[ \tan\alpha = \frac{6t}{2+8t^2} = \frac{3t}{1+4t^2} \]
Now,
\[ \tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta} {1-\tan\alpha\tan\beta} \]
\[ = \frac{\frac{3t}{1+4t^2}+t} {1-\frac{3t^2}{1+4t^2}} \]
\[ = \frac{4t(1+t^2)}{1+4t^2} \cdot \frac{1+4t^2}{1+t^2} \]
\[ =4t \]
Therefore,
\[ \tan(\alpha+\beta) = 4\tan\beta \]
Final Answer
\[ \boxed{\tan(\alpha+\beta)=4\tan\beta} \]
Hence, the correct option is (c) \(4\tan\beta\).