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I'm having trouble to understand the following equalities in these two equations, i.e. how to apply the addition formulas.

Firstly:

$$ \frac {1- \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} {1+ \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} = \frac {cos^2(\frac x2) - sin^2(\frac x2)} {cos^2(\frac x2) + sin^2(\frac x2)}= cos^2(\frac x2) - sin^2(\frac x2)$$

and secondly:

$$ \frac {2tan(\frac x2)} {1+tan^2(\frac x2)} = 2* \frac {sin(\frac x2)} {cos(\frac x2)(1+\frac {sin^2(\frac x2)}{cos^2(\frac x2)})} = 2* \frac {sin(\frac x2)cos(\frac x2)} {cos^2(\frac x2)+sin^2(\frac x2)}= 2sin(\frac x2)cos(\frac x2) $$

I don't understand all the steps in those equations unfortunately. Can someone help me out here?

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  • $\begingroup$ For first part $a=\sin^2\pi/2$, $b=\cos^2\pi/2$ ; $\Large\frac{1-\frac{a}{b}}{1+\frac{a}{b}}=\frac{\frac{b-a}{b}}{\frac{b+a}{b}}= \frac{b-a}{b+a}$ use $\sin^2\theta+\cos^2\theta=1$ $\endgroup$ – Vikram Oct 18 '14 at 10:50
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To get the first equality

$$\frac {1- \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} {1+ \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} \underbrace{=}_{(1)} \frac {cos^2(\frac x2) - sin^2(\frac x2)} {cos^2(\frac x2) + sin^2(\frac x2)}\underbrace{=}_{(2)} cos^2(\frac x2) - sin^2(\frac x2)$$

in step $(1)$ you multiply numerator and denominator by $\cos^2\frac x2$ (note that $\frac ab=\frac{ac}{bc}, \forall c\ne 0.$)

In step $(2)$ it is used the equality $\sin^2 t+\cos^2 t=1.$

To get the second equality

$$\frac {2tan(\frac x2)} {1+tan^2(\frac x2)} \underbrace{=}_{(1)} 2\frac {sin(\frac x2)} {cos(\frac x2)(1+\frac {sin^2(\frac x2)}{cos^2(\frac x2)})} \underbrace{=}_{(2)}2\frac {sin(\frac x2)cos(\frac x2)} {cos^2(\frac x2)+sin^2(\frac x2)}\underbrace{=}_{(3)} 2sin(\frac x2)cos(\frac x2)$$ note that:

$(1)$ is just the definition of tangent $\left(\tan t=\frac{\sin t}{\cos t}\right).$ In $(2)$ multiply numerator and denominator by $\cos \frac{x}{2}$ (note that, $\frac{a}{b}=\frac{ac}{bc}, \forall c\ne 0.$) In step $(3)$ it is used again the equality $\sin^2 t+\cos^2 t=1.$

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    $\begingroup$ In step (2) of the second identity, did you mean multiply numerator and denominator by $\cos \frac{x}{2}$? $\endgroup$ – N. F. Taussig Oct 18 '14 at 11:03
  • $\begingroup$ @N.F.Taussig Yes, you are right. Thank you. $\endgroup$ – mfl Oct 18 '14 at 11:04
  • $\begingroup$ Alright I got it, thanks :)! $\endgroup$ – holistic Oct 19 '14 at 14:51
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(2) that's multiply numerator and denominator by $\left(\cos\left(\tfrac x2\right)\right)^2$.

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