School exercises

Vereinfache und fasse soweit wie möglich zusammen:

$\frac{1 - (\cos (α))^{2}}{\sin (α) \cdot \sin (90 ° - α)} - \tan (α)$

For this task you need the following basic knowledge: Trigonometrische Funktionen
Gegeben:
$\frac{1 - (\cos (a))^{2}}{\sin (a) \cdot \sin (90 ° - a)} - \tan (a)$
Verwende den trigonometrischen Pythagoras:
$\sin^{2} (a) + \cos^{2} (a) = 1$
Damit kannst du den Zähler vereinfachen.
$= \frac{\sin^{2} (a)}{\sin (a) \cdot \sin (90 ° - a)} - \tan (a)$
Kürze im Bruch mit $\sin (a)$ .
$= \frac{\sin (a)}{\sin (90 ° - a)} - \tan (a)$
Wegen der Komplementbeziehung gilt:
$\sin (90 ° - a) = \cos (a)$ .
$= \frac{\sin (a)}{\cos (a)} - \tan (a)$
Benutze die Definition der Tangensfunktion.
$= \tan (a) - \tan (a) = 0$
$\frac{{(\cos (x + \frac{π}{2}))}^{2}}{1 - {(\frac{\sin (x)}{\tan (x)})}^{2}} + \frac{\sin (x) - \tan (x)}{\tan (x)}$

For this task you need the following basic knowledge: Trigonometrische Funktionen
Gegeben:
$\frac{{(\cos (x + \frac{π}{2}))}^{2}}{1 - {(\frac{\sin (x)}{\tan (x)})}^{2}} + \frac{\sin (x) - \tan (x)}{\tan (x)}$
Ziehe den hinteren Bruch auseinander.
$= \frac{{(\cos (x + \frac{π}{2}))}^{2}}{1 - {(\frac{\sin (x)}{\tan (x)})}^{2}} + \frac{\sin (x)}{\tan (x)} - \frac{\tan (x)}{\tan (x)}$
$= \frac{{(\cos (x + \frac{π}{2}))}^{2}}{1 - {(\frac{\sin (x)}{\tan (x)})}^{2}} + \frac{\sin (x)}{\tan (x)} - 1$
Aus der Formel für den Tangens folgt:
$\tan (x) = \frac{\sin (x)}{\cos (x)} | \cdot \cos (x)$
$\tan (x) \cdot \cos (x) = \sin (x) | : \tan (x)$
$\cos (x) = \frac{\sin (x)}{\tan (x)}$
Ersetze also $\frac{\sin (x)}{\tan (x)}$ durch $\cos (x)$
$= \frac{{(\cos (x + \frac{π}{2}))}^{2}}{1 - {(\frac{\sin (x)}{\tan (x)})}^{2}} + \cos (x) - 1$
Ersetze $\frac{\sin (x)}{\tan (x)}$ im Nenner des Bruches durch $\cos (x)$
$= \frac{{(\cos (x + \frac{π}{2}))}^{2}}{1 - (\cos (x))^{2}} + \cos (x) - 1$
Wenn du die Kosinusfunktion um $\frac{π}{2}$ verschiebt, also $\cos (x + \frac{π}{2})$ , erhältst du $- \sin (x)$ .
$= \frac{(- \sin (x))^{2}}{1 - (\cos (x))^{2}} + \cos (x) - 1$
Es gilt allgemein $(- x)^{2} = x^{2}$ , also auch $(- \sin (x))^{2} = (\sin (x))^{2}$ .
$= \frac{(\sin (x))^{2}}{1 - (\cos (x))^{2}} + \cos (x) - 1$
Verwende den trigometrischen Pythagoras:
$(\sin (x))^{2} + (\cos (x))^{2} = 1 | - (\cos (x))^{2}$
$(\sin (x))^{2} = 1 - (\cos (x))^{2}$
$= \frac{(\sin (x))^{2}}{(\sin (x))^{2}} + \cos (x) - 1$
$= 1 + \cos (x) - 1$
$= \cos (x)$

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