If \[ 2\tan\frac{\alpha}{2}=\tan\frac{\beta}{2}, \] prove that \[ \cos\alpha= \frac{3+5\cos\beta}{5+3\cos\beta} \]
Question
If \[ 2\tan\frac{\alpha}{2} = \tan\frac{\beta}{2}, \] prove that \[ \cos\alpha = \frac{3+5\cos\beta}{5+3\cos\beta}. \]
Solution
Given,
\[ 2\tan\frac{\alpha}{2} = \tan\frac{\beta}{2} \]
Therefore,
\[ \tan\frac{\alpha}{2} = \frac{1}{2}\tan\frac{\beta}{2} \]
Using the identity
\[ \cos\theta = \frac{1-\tan^2\frac{\theta}{2}} {1+\tan^2\frac{\theta}{2}} \]
For angle \(\alpha\),
\[ \cos\alpha = \frac{ 1-\tan^2\frac{\alpha}{2} }{ 1+\tan^2\frac{\alpha}{2} } \]
Substitute
\[ \tan\frac{\alpha}{2} = \frac{1}{2}\tan\frac{\beta}{2} \]
\[ \cos\alpha = \frac{ 1-\frac{1}{4}\tan^2\frac{\beta}{2} }{ 1+\frac{1}{4}\tan^2\frac{\beta}{2} } \]
Multiplying numerator and denominator by \(4\),
\[ \cos\alpha = \frac{ 4-\tan^2\frac{\beta}{2} }{ 4+\tan^2\frac{\beta}{2} } \]
Now use
\[ \tan^2\frac{\beta}{2} = \frac{1-\cos\beta}{1+\cos\beta} \]
Substituting,
\[ \cos\alpha = \frac{ 4-\frac{1-\cos\beta}{1+\cos\beta} }{ 4+\frac{1-\cos\beta}{1+\cos\beta} } \]
Taking LCM,
\[ = \frac{ \frac{ 4(1+\cos\beta)-(1-\cos\beta) }{ 1+\cos\beta } }{ \frac{ 4(1+\cos\beta)+(1-\cos\beta) }{ 1+\cos\beta } } \]
Simplifying numerator,
\[ 4+4\cos\beta-1+\cos\beta = 3+5\cos\beta \]
Simplifying denominator,
\[ 4+4\cos\beta+1-\cos\beta = 5+3\cos\beta \]
Therefore,
\[ \boxed{ \cos\alpha = \frac{3+5\cos\beta}{5+3\cos\beta} } \]
Hence proved.