Answer:
Se garantiza que [tex]\cos 2\alpha \cdot (\tan 2\alpha - \tan \alpha) = \tan \alpha[/tex].
Step-by-step explanation:
A continuación, vamos a emplear las siguientes identidades trigonométricas:
[tex]\cos 2\alpha = \cos^{2} \alpha - \sin^{2}\alpha[/tex] (1)
[tex]\tan 2\alpha = \frac{2\cdot \tan \alpha}{1-\tan^{2}\alpha}[/tex] (2)
Ahora aplicamos estas identidades a la identidad descrita en el enunciado:
[tex]\cos 2\alpha \cdot (\tan 2\alpha - \tan \alpha) = \tan \alpha[/tex]
[tex](\cos^{2}\alpha - \sin^{2}\alpha)\cdot \left(\frac{2\cdot \tan \alpha}{1-\tan^{2}\alpha} - \tan \alpha \right) = \tan \alpha[/tex]
[tex](\cos^{2}\alpha -\sin^{2}\alpha) \cdot \left[\frac{2- (1-\tan^{2}\alpha)}{1-\tan^{2}\alpha} \right]\cdot \tan\alpha = \tan \alpha[/tex]
[tex](\cos^{2}\alpha - \sin^{2}\alpha) \cdot \left[\frac{1+\tan^{2}\alpha}{1-\tan^{2}\alpha} \right] = 1[/tex]
[tex](\cos^{2}\alpha -\sin^{2}\alpha)\cdot (1 + \tan^{2}\alpha) = 1 - \tan^{2}\alpha[/tex]
[tex]\cos^{2}\alpha \cdot (1-\tan^{2}\alpha)\cdot (1 + \tan^{2}\alpha) = (1-\tan^{2}\alpha)[/tex]
[tex]\cos^{2}\alpha \cdot (1+\tan^{2}\alpha) = 1[/tex]
[tex]\cos^{2}\alpha\cdot \sec^{2}\alpha = 1[/tex]
[tex]1 = 1[/tex]
Por tanto, se garantiza que [tex]\cos 2\alpha \cdot (\tan 2\alpha - \tan \alpha) = \tan \alpha[/tex].
