If \[ \cos x= \frac{\cos\alpha+\cos\beta} {1+\cos\alpha\cos\beta}, \] prove that \[ \tan\frac{x}{2} = \pm \tan\frac{\alpha}{2} \tan\frac{\beta}{2} \]
Question
If \[ \cos x= \frac{\cos\alpha+\cos\beta} {1+\cos\alpha\cos\beta}, \] prove that \[ \tan\frac{x}{2} = \pm \tan\frac{\alpha}{2} \tan\frac{\beta}{2}. \]
Solution
Using the identity
\[ \tan^2\frac{x}{2} = \frac{1-\cos x}{1+\cos x} \]
Substitute the given value of \[ \cos x \]
\[ \tan^2\frac{x}{2} = \frac{ 1- \frac{\cos\alpha+\cos\beta} {1+\cos\alpha\cos\beta} } { 1+ \frac{\cos\alpha+\cos\beta} {1+\cos\alpha\cos\beta} } \]
Taking LCM in numerator and denominator,
\[ = \frac{ \frac{ 1+\cos\alpha\cos\beta-\cos\alpha-\cos\beta } { 1+\cos\alpha\cos\beta } } { \frac{ 1+\cos\alpha\cos\beta+\cos\alpha+\cos\beta } { 1+\cos\alpha\cos\beta } } \]
\[ = \frac{ 1+\cos\alpha\cos\beta-\cos\alpha-\cos\beta } { 1+\cos\alpha\cos\beta+\cos\alpha+\cos\beta } \]
Factor numerator:
\[ 1+\cos\alpha\cos\beta-\cos\alpha-\cos\beta = (1-\cos\alpha)(1-\cos\beta) \]
Factor denominator:
\[ 1+\cos\alpha\cos\beta+\cos\alpha+\cos\beta = (1+\cos\alpha)(1+\cos\beta) \]
Therefore,
\[ \tan^2\frac{x}{2} = \frac{ (1-\cos\alpha)(1-\cos\beta) }{ (1+\cos\alpha)(1+\cos\beta) } \]
Using
\[ \tan^2\frac{\theta}{2} = \frac{1-\cos\theta}{1+\cos\theta} \]
we get
\[ \tan^2\frac{x}{2} = \tan^2\frac{\alpha}{2} \tan^2\frac{\beta}{2} \]
Taking square root on both sides,
\[ \tan\frac{x}{2} = \pm \tan\frac{\alpha}{2} \tan\frac{\beta}{2} \]
Hence proved.
Final Answer
\[ \boxed{ \tan\frac{x}{2} = \pm \tan\frac{\alpha}{2} \tan\frac{\beta}{2} } \]