Question

$demonstre \: que \: \frac{1 - cos \: x}{1 + cos \: x} = {tg}^{2} \frac{x}{2}$
​

Answer 1

Do cosseno do arco duplo temos:

$\mathsf{cos(2x)=2cos^2(x)-1}$

fazendo

$\mathsf{t=2x\to~x=\dfrac{t}{2}}$ temos:

$\mathsf{cos(t)=2cos^2(\dfrac{t}{2})-1}$

Daí

$\mathsf{cos(x)=2cos^2(\dfrac{x}{2})-1}$

Vamos usar essa identidade para fazer a demonstração pedida.

$\mathsf{\dfrac{1-cos(x)}{1+cos(x)}}\\\mathsf{\dfrac{1-\left[2cos^2(\dfrac{x}{2})-1\right]}{1+2cos^2(\dfrac{x}{2})-1}}$

$\mathsf{\dfrac{1-2cos^2(\dfrac{x}{2})+1}{2cos^2(\dfrac{x}{2})}}\\\mathsf{\dfrac{2-2cos^2(\dfrac{x}{2})}{2cos^2(\dfrac{x}{2})}}$

$\mathsf{\dfrac{2(1-cos^2(\dfrac{x}{2}))}{2cos^2(\dfrac{x}{2})}}\\\mathsf{\dfrac{1-cos^2(\dfrac{x}{2})}{cos^2(\dfrac{x}{2})}}$

$\mathsf{\dfrac{sen^2(\dfrac{x}{2})}{cos^2(\dfrac{x}{2})}} = \large\boxed{\boxed{\boxed{\boxed{\mathsf{tg^2(\dfrac{x}{2})}}}}}$