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Demonstração:

A demonstração abaixo parte das seguintes identidades:

[tex]sin\left(2x\right) = 2 \cdot sin\left(x\right) \cdot cos\left(x\right), \forall \,\, x \in \mathbb{R};\\\\\\cos\left(2x\right) = cos^2\left(x\right) - sin^2\left(x\right), \forall \,\, x \in \mathbb{R}.[/tex]

A partir das identidades acima, deduziremos as expressões de [tex]tg\left(2x\right)[/tex] e [tex]sec\left(2x\right).[/tex]

Para todo [tex]x \in \mathbb{R}\, | \, x \neq \dfrac{\pi}{4} + \dfrac{k\pi}{2}, k \in \mathbb{Z},[/tex] temos:

[tex]tg\left(2x\right) = \dfrac{\sin\left(2x\right)}{\cos\left(2x\right)}\\\\\\= \dfrac{2 \cdot sin\left(x\right) \cdot cos \left(x\right)}{cos^2\left(x\right) - sin^2\left(x\right)}\\\\\\= \dfrac{\dfrac{2 \cdot sin\left(x\right) \cdot cos \left(x\right)}{cos^2\left(x\right)}}{\dfrac{cos^2\left(x\right) - sin^2\left(x\right)}{cos^2\left(x\right)}}\\\\\\= \boxed{\dfrac{2 \cdot tg \left(x\right)}{1 - tg^2\left(x\right)}}[/tex]

Para todo [tex]x \in \mathbb{R}\, | \, x \neq \dfrac{\pi}{4} + \dfrac{k\pi}{2}, k \in \mathbb{Z},[/tex] temos:

[tex]sec\left(2x\right) = \dfrac{1}{cos\left(2x\right)}\\\\\\= \dfrac{1}{cos^2\left(x\right)-sin^2\left(x\right)}[/tex]

Aplicando a identidade trigonométrica fundamental, podemos substituir a unidade no numerador pela expressão [tex]cos^2\left(x\right) + sin^2\left(x\right):[/tex]

[tex]= \dfrac{cos^2\left(x\right) + sin^2\left(x\right) }{cos^2\left(x\right)-sin^2\left(x\right)}\\\\\\= \dfrac{\dfrac{cos^2\left(x\right) + sin^2\left(x\right)}{cos^2\left(x\right)}}{\dfrac{cos^2\left(x\right)-sin^2\left(x\right)}{cos^2\left(x\right)}}\\\\\\= \boxed{\dfrac{1 + tg^2\left(x\right)}{1 - tg^2\left(x\right)}}[/tex]

Observemos as duas condições para que a expressão [tex]\dfrac{tg\left(2x\right)}{sec\left(2x\right)+1}[/tex] esteja definida em [tex]\mathbb{R}:[/tex]

[tex]tg\left(2x\right) \in \mathbb{R} \Longleftrightarrow cos\left(2x\right) \neq 0 \\\\\Longleftrightarrow 2x \neq \dfrac{\pi}{2} + k\pi, k \in \mathbb{Z} \\\\\Longleftrightarrow x \neq \dfrac{\pi}{4} + \dfrac{k\pi}{2}, k \in \mathbb{Z}[/tex]

[tex]sec\left(2x\right) + 1 \neq 0\\\\\Longleftrightarrow sec\left(2x\right) \neq -1\\\\\Longleftrightarrow cos\left(2x\right) \neq -1\\\\\Longleftrightarrow 2x \neq \pi + 2k\pi, k \in \mathbb{Z}\\\\\Longleftrightarrow x \neq \dfrac{\pi}{2} + k\pi, k \in \mathbb{Z}.[/tex]

Assim, para todo [tex]x \in \mathbb{R} \, | \, x \neq \dfrac{\pi}{4} + \dfrac{k\pi}{2} \,\,\,e\,\,\,x \neq \dfrac{\pi}{2} + k\pi, k \in \mathbb{Z},[/tex] temos:

[tex]\dfrac{tg\left(2x\right)}{sec\left(2x\right)+1} = \dfrac{\dfrac{2 \cdot tg\left(x\right)}{1-tg^2\left(x\right)}}{\dfrac{1+tg^2\left(x\right)}{1- tg^2\left(x\right)} + 1}\\\\\\= \dfrac{\dfrac{2 \cdot tg\left(x\right)}{1-tg^2\left(x\right)}}{\dfrac{1+tg^2\left(x\right)+1-tg^2\left(x\right)}{1- tg^2\left(x\right)}}\\\\\\= \dfrac{2 \cdot tg\left(x\right)}{2}\\\\\\= tg\left(x\right).[/tex]

Q.E.D.