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Does anyone know how the equation in red was derived?

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    Write $t = \cos(\theta /4) / \sin(\theta /4)$ and simplify. – 2017-01-19

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First part:

$$\cos\theta=\cos^2\frac\theta2-\sin^2\frac\theta2=\left(\cos^2\frac\theta4-\sin^2\frac\theta4\right)^2-4\sin^2\frac\theta4\cos^2\frac\theta4.$$

Second part:

$$\cos^2\frac\theta4=\frac1{1+\tan^2\frac\theta4}=\frac1{1+t^2},\\ \sin^2\frac\theta4=\tan^2\frac\theta4\cos^2\frac\theta4=\frac{t^2}{1+t^2}.$$

But $t=\tan\frac\theta4$, not $\cot\frac\theta4$ nor $\arctan\frac\theta4$ !

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    I got this question from a book. "Differential Geometry - Martin Lipschultz" – 2017-01-19
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    @Isaac: obviously. But that doesn't make $t=\text{Tan}^{-1}(\theta/4)$. – 2017-01-19