Question

If $cos\theta =\frac{cos\alpha +cos\beta }{1+cos\alpha cos\beta }$, prove that $tan\frac{\theta }{2}=\pm tan\frac{\alpha }{2}tan\frac{\beta }{2}$

Answer 1

We have,

$cos\theta =\frac{cos\alpha +cos\beta }{1+cos\alpha cos\beta }$

Now,

$cos\theta =\frac{1-ta{n}^2\frac{\theta }{2}}{1+ta{n}^2\frac{\theta }{2}}\Rightarrow \frac{1-ta{n}^2\frac{\theta }{2}}{1+ta{n}^2\frac{\theta }{2}}=\frac{cos\alpha +cos\beta }{1+cos\alpha cos\beta }$

by componende and dividendo, we get

$\frac{(1-ta{n}^2\frac{\theta }{2})+(1+ta{n}^2\frac{\theta }{2})}{(1-ta{n}^2\frac{\theta }{2})-(1+ta{n}^2\frac{\theta }{2})}=\frac{1+cos\alpha cos\beta +cos\alpha +cos\beta }{-(1+cos\alpha cos\beta -cos\alpha -cos\beta )}\Rightarrow \frac{2}{2ta{n}^2\frac{\theta }{2}}=\frac{(1+cos\alpha )(1+cos\beta )}{(1-cos\alpha )(1-cos\beta )}\Rightarrow ta{n}^2\frac{\theta }{2}=\frac{(1-cos\alpha )(1-cos\beta )}{(1+cos\alpha )(1+cos\beta )}$

$=\frac{2si{n}^2\frac{\alpha }{2}.2si{n}^2\frac{\beta }{2}}{2co{s}^2\frac{\alpha }{2}.2co{s}^2\frac{\beta }{2}}\Rightarrow tan\frac{\theta }{2}=\pm tan\frac{\theta }{2}.tan\frac{\beta }{2}$