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Beliebt Trigonometrie >

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

Lösung

beweisen

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

Lösung

Wahr

Schritte zur Lösung

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

Manipuliere die linke Seite

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Verwende die Pythagoreische Identität:

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

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

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

Vereinfache

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

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

Faktorisiere

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

sec^{4} (x) - 1

Schreibe

sec^{4} (x) - 1

um:

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

sec^{4} (x) - 1

Schreibe

1

um:

1^{2}

= sec^{4} (x) - 1^{2}

Wende Exponentenregel an:

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

sec^{4} (x) = (sec^{2} (x))^{2}

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

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

Wende Formel zur Differenz von zwei Quadraten an:

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

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

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

Faktorisiere

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

sec^{2} (x) - 1

Schreibe

1

um:

1^{2}

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

Wende Formel zur Differenz von zwei Quadraten an:

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

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

= (sec (x) + 1) (sec (x) - 1)

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

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

Faktorisiere

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

sec^{2} (x) - 1

Schreibe

1

um:

1^{2}

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

Wende Formel zur Differenz von zwei Quadraten an:

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

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

= (sec (x) + 1) (sec (x) - 1)

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

Streiche

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

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

Streiche die gemeinsamen Faktoren:

sec (x) + 1

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

Streiche die gemeinsamen Faktoren:

sec (x) - 1

= sec^{2} (x) + 1

= sec^{2} (x) + 1

= sec^{2} (x) + 1

Verwende die Pythagoreische Identität:

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

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

Vereinfache

= tan^{2} (x) + 2

= tan^{2} (x) + 2

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

\Rightarrow Wahr

Beliebte Beispiele

beweisen tan^2(x)-sec^2(x)=-1

p ro v e tan^{2} (x) - sec^{2} (x) = - 1

beweisen cos(-θ)=cos(θ)

p ro v e cos (- θ) = cos (θ)

beweisen 1+sin(2θ)=(sin(θ)+cos(θ))^2

p ro v e 1 + sin (2 θ) = (sin (θ) + cos (θ))^{2}

beweisen cos((5pi)/4-x)=-(sqrt(2))/2 (cos(x)+sin(x))

p ro v e cos (\frac{5 π}{4} - x) = - \frac{2}{2} (cos (x) + sin (x))

beweisen 5cos^2(θ)+3sin^2(θ)=3+2cos^2(θ)

p ro v e 5 cos^{2} (θ) + 3 sin^{2} (θ) = 3 + 2 cos^{2} (θ)