Prove that:
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We can differentiate each term and show that for all . Thus and taking give . There are many calculations but it isn't difficult for all that.
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About LinkBacks![2\arctan\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right] = \arccos\left[\frac{b + a\cos x}{a + b\cos x}\right]](http://latex.codecogs.com/png.latex?2\arctan\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right] = \arccos\left[\frac{b + a\cos x}{a + b\cos x}\right])
