Is csch^2(x)=coth^2(x)-1 ?

The answer to whether csch^2(x)=coth^2(x)-1 is True

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prove csch^2(x)=coth^2(x)-1

prove

csch^{2} (x) = coth^{2} (x) - 1

True

Solution steps

csch^{2} (x) = coth^{2} (x) - 1

Manipulating right side

coth^{2} (x) - 1

Rewrite using trig identities

coth^{2} (x) - 1

Use the Hyperbolic identity:

coth (x) = \frac{cosh ( x )}{sinh ( x )}

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

Rewrite

1

\frac{sinh ^{2} ( x )}{sinh ^{2} ( x )}

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

Simplify

(\frac{cosh ( x )}{sinh ( x )})^{2} - \frac{sinh ^{2} ( x )}{sinh ^{2} ( x )} : \frac{cosh ^{2} ( x ) - sinh ^{2} ( x )}{sinh ^{2} ( x )}

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

Apply rule

\frac{a}{a} = 1

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

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

Apply exponent rule:

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sinh^{2} (x) = sinh^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Hyperbolic identity:

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

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

Use the Hyperbolic identity:

sinh (x) = \frac{1}{csch ( x )}

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

Simplify

\frac{1}{( \frac{1}{csch ( x )} ) ^{2}} : csch^{2} (x)

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= csch^{2} (x)

= csch^{2} (x)

= csch^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True