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العربية

beweisen 1/((1-\frac{1){(1-1/((1-cosh^2(x))))})}=cosh^2(x)

Lösung

beweisen

\frac{1}{1 - \frac{1}{( 1 - \frac{1}{( 1 - c o s h ^{2} ( x ) )} )}} = cosh^{2} (x)

Wahr

Schritte zur Lösung

\frac{1}{1 - \frac{1}{1 - \frac{1}{1 - c o s h ^{2} ( x )}}} = cosh^{2} (x)

Manipuliere die linke Seite

\frac{1}{1 - \frac{1}{1 - \frac{1}{1 - c o s h ^{2} ( x )}}}

Vereinfache

\frac{1}{1 - \frac{1}{1 - \frac{1}{1 - c o s h ^{2} ( x )}}} : cosh^{2} (x)

\frac{1}{1 - \frac{1}{1 - \frac{1}{1 - c o s h ^{2} ( x )}}}

\frac{1}{1 - \frac{1}{1 - c o s h ^{2} ( x )}} = - \frac{1 - cosh ^{2} ( x )}{cosh ^{2} ( x )}

\frac{1}{1 - \frac{1}{1 - c o s h ^{2} ( x )}}

Füge

1 - \frac{1}{1 - cosh ^{2} ( x )}

zusammen:

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

1 - \frac{1}{1 - cosh ^{2} ( x )}

Wandle das Element in einen Bruch um:

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

= \frac{1 \cdot ( 1 - cosh ^{2} ( x ) )}{1 - cosh ^{2} ( x )} - \frac{1}{1 - cosh ^{2} ( x )}

Da die Nenner gleich sind, fasse die Brüche zusammen.:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot ( 1 - cosh ^{2} ( x ) ) - 1}{1 - cosh ^{2} ( x )}

1 \cdot (1 - cosh^{2} (x)) - 1 = - cosh^{2} (x)

1 \cdot (1 - cosh^{2} (x)) - 1

1 \cdot (1 - cosh^{2} (x)) = 1 - cosh^{2} (x)

1 \cdot (1 - cosh^{2} (x))

Multipliziere:

1 \cdot (1 - cosh^{2} (x)) = (1 - cosh^{2} (x))

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

Entferne die Klammern:

(a) = a

= 1 - cosh^{2} (x)

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

Fasse gleiche Terme zusammen

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

1 - 1 = 0

= - cosh^{2} (x)

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

Wende Bruchregel an:

\frac{- a}{b} = - \frac{a}{b}

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

= \frac{1}{- \frac{c o s h ^{2} ( x )}{- c o s h ^{2} ( x ) + 1}}

Wende Bruchregel an:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{1}{\frac{c o s h ^{2} ( x )}{1 - c o s h ^{2} ( x )}}

Wende Bruchregel an:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

\frac{1}{\frac{c o s h ^{2} ( x )}{1 - c o s h ^{2} ( x )}} = \frac{1 - cosh ^{2} ( x )}{cosh ^{2} ( x )}

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

= \frac{1}{1 - ( - \frac{- c o s h ^{2} ( x ) + 1}{c o s h ^{2} ( x )} )}

Wende Regel an

- (- a) = a

= \frac{1}{1 + \frac{1 - c o s h ^{2} ( x )}{c o s h ^{2} ( x )}}

Füge

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

zusammen:

\frac{1}{cosh ^{2} ( x )}

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

Wandle das Element in einen Bruch um:

1 = \frac{1 cosh ^{2} ( x )}{cosh ^{2} ( x )}

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

Da die Nenner gleich sind, fasse die Brüche zusammen.:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

1 \cdot cosh^{2} (x) + 1 - cosh^{2} (x) = 1

1 \cdot cosh^{2} (x) + 1 - cosh^{2} (x)

Fasse gleiche Terme zusammen

= 1 \cdot cosh^{2} (x) - cosh^{2} (x) + 1

Addiere gleiche Elemente:

1 \cdot cosh^{2} (x) - cosh^{2} (x) = 0

= 1

= \frac{1}{cosh ^{2} ( x )}

= \frac{1}{\frac{1}{c o s h ^{2} ( x )}}

Wende Bruchregel an:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{cosh ^{2} ( x )}{1}

Wende Regel an

\frac{a}{1} = a

= cosh^{2} (x)

= cosh^{2} (x)

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

\Rightarrow Wahr

p ro v e (\frac{sec ( θ )}{cos ( θ )}) - (\frac{tan ( θ )}{cot ( θ )}) = 1

p ro v e sin (π) = sin (3 π)

p ro v e \frac{2 + csc ( x )}{sec ( x )} - 2 cos (x) = cot (x)

p ro v e \frac{cos ( 6 0 ^{\circ} )}{sin ( 6 0 ^{\circ} )} = cot (6 0^{\circ})

p ro v e \frac{1}{sec ( x ) ( sec ( x ) - cos ( x ) )} = cot^{2} (x)