If 2 tan α = 3 tan β, prove that \[ \tan(\alpha-\beta)=\frac{\sin2\beta}{5-\cos2\beta} \]
Question
If \[ 2\tan\alpha = 3\tan\beta, \] prove that \[ \tan(\alpha-\beta) = \frac{\sin2\beta}{5-\cos2\beta}. \]
Solution
Given,
\[ 2\tan\alpha = 3\tan\beta \]
Therefore,
\[ \tan\alpha = \frac{3}{2}\tan\beta \]
Using the identity
\[ \tan(\alpha-\beta) = \frac{\tan\alpha-\tan\beta} {1+\tan\alpha\tan\beta} \]
Substitute \[ \tan\alpha=\frac{3}{2}\tan\beta \]
\[ \tan(\alpha-\beta) = \frac{ \frac{3}{2}\tan\beta-\tan\beta } { 1+\frac{3}{2}\tan^2\beta } \]
\[ = \frac{ \frac{1}{2}\tan\beta } { 1+\frac{3}{2}\tan^2\beta } \]
\[ = \frac{ \tan\beta } { 2+3\tan^2\beta } \]
Now express everything in terms of \[ \sin2\beta \quad \text{and} \quad \cos2\beta \]
Using
\[ \tan\beta = \frac{\sin2\beta}{1+\cos2\beta} \]
and
\[ \tan^2\beta = \frac{1-\cos2\beta}{1+\cos2\beta} \]
Substituting,
\[ \tan(\alpha-\beta) = \frac{ \frac{\sin2\beta}{1+\cos2\beta} } { 2+3\left( \frac{1-\cos2\beta}{1+\cos2\beta} \right) } \]
Take LCM in the denominator:
\[ = \frac{ \frac{\sin2\beta}{1+\cos2\beta} } { \frac{ 2(1+\cos2\beta)+3(1-\cos2\beta) } {1+\cos2\beta} } \]
\[ = \frac{ \sin2\beta } { 2+2\cos2\beta+3-3\cos2\beta } \]
\[ = \frac{ \sin2\beta } { 5-\cos2\beta } \]
Hence proved.
Final Answer
\[ \boxed{ \tan(\alpha-\beta) = \frac{\sin2\beta}{5-\cos2\beta} } \]