Respondido

Demonstre a identidade trigonométrica cos²x - sen²x = 1-tg x
                                                         --------------------   --------
                                                          1 + tg x            sec² x
Dica : Lembre-se que a² - b² = (a + b).(a - b) .




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Resposta :

MATHSPHIS
[tex]\frac{cos^2x-sen^2x}{1+tgx}=\frac{1-tgx}{sec^2x} \\ \\ \frac{cos^2x-sen^2x}{1+\frac{senx}{cosx}}=\frac{1-\frac{senx}{cosx}}{\frac{1}{cos^2x}} \\ \\ \frac{cos^2x-sen^2x}{\frac{cosx+senx}{cosx}}=\frac{\frac{cosx-senx}{cosx}}{\frac{1}{cos^2x}} \\ \\ (cosx+senx)(cosx-senx).\frac{cosx}{cosx+senx}=\frac{cosx-senx}{cosx}.cos^2x \\ \\ \boxed{(cosx-senx)cosx=(cosx-senx).cosx}[/tex]

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