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beweisen (sec^4(α)-1)/(tan^2(α))=sec^2(α)+1

Lösung

beweisen

\frac{sec ^{4} ( α ) - 1}{tan ^{2} ( α )} = sec^{2} (α) + 1

Wahr

Schritte zur Lösung

\frac{sec ^{4} ( α ) - 1}{tan ^{2} ( α )} = sec^{2} (α) + 1

Manipuliere die linke Seite

\frac{sec ^{4} ( α ) - 1}{tan ^{2} ( α )}

Umschreiben mit Hilfe von Trigonometrie-Identitäten

\frac{sec ^{4} ( α ) - 1}{tan ^{2} ( α )}

Verwende die Pythagoreische Identität:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= \frac{sec ^{4} ( α ) - 1}{sec ^{2} ( α ) - 1}

Vereinfache

\frac{sec ^{4} ( α ) - 1}{sec ^{2} ( α ) - 1} : sec^{2} (α) + 1

\frac{sec ^{4} ( α ) - 1}{sec ^{2} ( α ) - 1}

Faktorisiere

sec^{4} (α) - 1 : (sec^{2} (α) + 1) (sec (α) + 1) (sec (α) - 1)

sec^{4} (α) - 1

Schreibe

sec^{4} (α) - 1

um:

(sec^{2} (α))^{2} - 1^{2}

sec^{4} (α) - 1

Schreibe

1

um:

1^{2}

= sec^{4} (α) - 1^{2}

Wende Exponentenregel an:

a^{b c} = (a^{b})^{c}

sec^{4} (α) = (sec^{2} (α))^{2}

= (sec^{2} (α))^{2} - 1^{2}

= (sec^{2} (α))^{2} - 1^{2}

Wende Formel zur Differenz von zwei Quadraten an:

x^{2} - y^{2} = (x + y) (x - y)

(sec^{2} (α))^{2} - 1^{2} = (sec^{2} (α) + 1) (sec^{2} (α) - 1)

= (sec^{2} (α) + 1) (sec^{2} (α) - 1)

Faktorisiere

sec^{2} (α) - 1 : (sec (α) + 1) (sec (α) - 1)

sec^{2} (α) - 1

Schreibe

1

um:

1^{2}

= sec^{2} (α) - 1^{2}

Wende Formel zur Differenz von zwei Quadraten an:

x^{2} - y^{2} = (x + y) (x - y)

sec^{2} (α) - 1^{2} = (sec (α) + 1) (sec (α) - 1)

= (sec (α) + 1) (sec (α) - 1)

= (sec^{2} (α) + 1) (sec (α) + 1) (sec (α) - 1)

= \frac{( sec ^{2} ( α ) + 1 ) ( sec ( α ) + 1 ) ( sec ( α ) - 1 )}{sec ^{2} ( α ) - 1}

Faktorisiere

sec^{2} (α) - 1 : (sec (α) + 1) (sec (α) - 1)

sec^{2} (α) - 1

Schreibe

1

um:

1^{2}

= sec^{2} (α) - 1^{2}

Wende Formel zur Differenz von zwei Quadraten an:

x^{2} - y^{2} = (x + y) (x - y)

sec^{2} (α) - 1^{2} = (sec (α) + 1) (sec (α) - 1)

= (sec (α) + 1) (sec (α) - 1)

= \frac{( sec ^{2} ( α ) + 1 ) ( sec ( α ) + 1 ) ( sec ( α ) - 1 )}{( sec ( α ) + 1 ) ( sec ( α ) - 1 )}

Streiche

\frac{( sec ^{2} ( α ) + 1 ) ( sec ( α ) + 1 ) ( sec ( α ) - 1 )}{( sec ( α ) + 1 ) ( sec ( α ) - 1 )} : sec^{2} (α) + 1

\frac{( sec ^{2} ( α ) + 1 ) ( sec ( α ) + 1 ) ( sec ( α ) - 1 )}{( sec ( α ) + 1 ) ( sec ( α ) - 1 )}

Streiche die gemeinsamen Faktoren:

sec (α) + 1

= \frac{( sec ^{2} ( α ) + 1 ) ( sec ( α ) - 1 )}{sec ( α ) - 1}

Streiche die gemeinsamen Faktoren:

sec (α) - 1

= sec^{2} (α) + 1

= sec^{2} (α) + 1

= sec^{2} (α) + 1

= sec^{2} (α) + 1

Wir haben gezeigt, dass beide Seiten die gleiche Form annehmen können

\Rightarrow Wahr

p ro v e 2 sin (x) = \frac{4 cos ( x ) - 1}{tan ( x )}

p ro v e cos (8 x) = cos^{2} (4 x) - sin^{2} (4 x)

p ro v e tan (\frac{π}{2} - x) cot (x) = csc^{2} (x) - 1

p ro v e cot^{2} (α) = cos^{2} (α) + (cot (α) \cdot cos (α))^{2}

p ro v e \frac{1}{tan ( α )} \cdot \frac{1}{cos ( α )} = \frac{1}{sin ( α )}