What is the value of coth(1) ?

The value of coth(1) is (e^2+1)/(e^2-1)

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coth(1)

coth (1)

\frac{e ^{2} + 1}{e ^{2} - 1}

Decimal Notation

1.31303 \dots

Solution steps

coth (1)

Use the Hyperbolic identity:

coth (x) = \frac{e ^{x} + e ^{- x}}{e ^{x} - e ^{- x}}

= \frac{e ^{1} + e ^{- 1}}{e ^{1} - e ^{- 1}}

\frac{e ^{1} + e ^{- 1}}{e ^{1} - e ^{- 1}} = \frac{e ^{2} + 1}{e ^{2} - 1}

\frac{e ^{1} + e ^{- 1}}{e ^{1} - e ^{- 1}}

Apply rule

a^{1} = a

e^{1} = e

= \frac{e + e ^{- 1}}{e - e ^{- 1}}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

e^{- 1} = \frac{1}{e}

= \frac{e + \frac{1}{e}}{e - \frac{1}{e}}

Join

e - \frac{1}{e} : \frac{e ^{2} - 1}{e}

e - \frac{1}{e}

Convert element to fraction:

e = \frac{ee}{e}

= \frac{ee}{e} - \frac{1}{e}

Since the denominators are equal, combine the fractions:

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

= \frac{ee - 1}{e}

ee - 1 = e^{2} - 1

ee - 1

ee = e^{2}

ee

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

ee = e^{1 + 1}

= e^{1 + 1}

Add the numbers:

1 + 1 = 2

= e^{2}

= e^{2} - 1

= \frac{e ^{2} - 1}{e}

= \frac{e + \frac{1}{e}}{\frac{e ^{2} - 1}{e}}

Join

e + \frac{1}{e} : \frac{e ^{2} + 1}{e}

e + \frac{1}{e}

Convert element to fraction:

e = \frac{ee}{e}

= \frac{ee}{e} + \frac{1}{e}

Since the denominators are equal, combine the fractions:

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

= \frac{ee + 1}{e}

ee + 1 = e^{2} + 1

ee + 1

ee = e^{2}

ee

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

ee = e^{1 + 1}

= e^{1 + 1}

Add the numbers:

1 + 1 = 2

= e^{2}

= e^{2} + 1

= \frac{e ^{2} + 1}{e}

= \frac{\frac{e ^{2} + 1}{e}}{\frac{e ^{2} - 1}{e}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

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

Cancel the common factor:

e

= \frac{e ^{2} + 1}{e ^{2} - 1}

= \frac{e ^{2} + 1}{e ^{2} - 1}

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