Question

Show that:

${\cos }^{-1}\left[\frac{\cos \alpha +\cos \beta }{1+\cos \alpha \cos \beta }\right]=2{\tan }^{-1}\left(\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\right)$

Answer 1

${\cos }^{-1}\left[\frac{\cos \alpha +\cos \beta }{1+\cos \alpha \cos \beta }\right]=2{\tan }^{-1}\left(\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\right)$

RHS $=2ta{n}_{-1}\left[tan\frac{\alpha }{2}tan\frac{\beta }{2}\right]$

$=co{s}_{-1}\left[\frac{1-ta{n}^2\alpha /2ta{n}^2\beta /2}{1+ta{n}^2\alpha /2ta{n}^2\beta /2}\right]$

$=co{s}^{-1}\left[\frac{\frac{1-si{n}^2\alpha /2si{n}^2\beta /2}{co{s}^2\alpha /2co{s}^2\beta /2}}{1+\frac{si{n}^2\alpha /2si{n}^2\beta /2}{co{s}^2\alpha /2co{s}^2\beta /2}}\right]$

$=co{s}^{-1}\left[\frac{co{s}^2\alpha /2co{s}^2\beta /2-si{n}^2\alpha /2si{n}^2\beta /2}{co{s}^2\alpha /2co{s}^2\beta /2+si{n}^2\alpha /2si{n}^2\beta /2}\right]$

$=co{s}^{-1}\left[\frac{2co{s}^2\alpha /2co{s}^2\beta /2-si{n}^2\alpha /2si{n}^2\beta /2}{2co{s}^2\alpha /2co{s}^2\beta /2+2si{n}^2\alpha /2si{n}^2\beta /2}\right]$

$=co{s}^{-1}\left[\frac{(1+cos\alpha )(1+cos\beta )-(1-cos\alpha )(1-cos\beta )}{(1+cos\alpha )(1+cos\beta )+(1-cos\alpha )(1-cos\beta )}\right]$

$=co{s}^{-1}\left[\frac{cos\alpha +cos\beta }{1-cos\alpha cos\beta }\right]$

$=LHS$