If \(2\tan\alpha = 3\tan\beta\), Find \( \tan(\alpha-\beta) \)
Question
If
\[ 2\tan\alpha = 3\tan\beta \]
then
\[ \tan(\alpha-\beta) \]
is equal to
(a) \(\dfrac{\sin2\beta}{5-\cos2\beta}\)
(b) \(\dfrac{\cos2\beta}{5-\cos2\beta}\)
(c) \(\dfrac{\sin2\beta}{5+\cos2\beta}\)
(d) none of these
Solution
Given,
\[ 2\tan\alpha=3\tan\beta \]
\[ \tan\alpha=\frac{3}{2}\tan\beta \]
Using the identity
\[ \tan(\alpha-\beta) = \frac{\tan\alpha-\tan\beta} {1+\tan\alpha\tan\beta} \]
Substitute \(\tan\alpha=\frac32\tan\beta\):
\[ \tan(\alpha-\beta) = \frac{\frac32\tan\beta-\tan\beta} {1+\frac32\tan^2\beta} \]
\[ = \frac{\tan\beta} {2+3\tan^2\beta} \]
Multiply numerator and denominator by \(\cos^2\beta\):
\[ = \frac{\sin\beta\cos\beta} {2\cos^2\beta+3\sin^2\beta} \]
Using
\[ \sin\beta\cos\beta=\frac{\sin2\beta}{2} \]
\[ 2\cos^2\beta+3\sin^2\beta = \frac{5-\cos2\beta}{2} \]
Therefore,
\[ \tan(\alpha-\beta) = \frac{\frac{\sin2\beta}{2}} {\frac{5-\cos2\beta}{2}} \]
\[ = \frac{\sin2\beta}{5-\cos2\beta} \]
Final Answer
\[ \boxed{\tan(\alpha-\beta)=\frac{\sin2\beta}{5-\cos2\beta}} \]
Hence, the correct option is (a).