Question

Prove the identity.

17. \({\coth ^2}x - 1 = {{\mathop{\rm csch}\nolimits} ^2}x\)

Short answer

It is proved that \({\coth ^2}x - 1 = {{\mathop{\rm csch}\nolimits} ^2}x\).

Step-by-step solution

Definition of Hyperbolic function

The formulas of the hyperbolic function as shown below:

\(\begin{aligned}\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\rm csch}\nolimits} x = \frac{1}{{\sinh x}}\\\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\rm sech}\nolimits} x = \frac{1}{{\cosh x}}\\\tanh x = \frac{{\sinh x}}{{\cosh x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\coth x = \frac{{\cosh x}}{{\sinh x}}\end{aligned}\)

Prove the identity\({\coth ^2}x - 1 = {{\mathop{\rm csch}\nolimits} ^2}x\)

Divide each side of the identity \({\cosh ^2}x - {\sinh ^2}x = 1\) by \({\sinh ^2}x\)as shown below:

\(\begin{aligned}\frac{{{{\cosh }^2}x}}{{{{\sinh }^2}x}} - \frac{{{{\sinh }^2}x}}{{{{\sinh }^2}x}} &= \frac{1}{{{{\sinh }^2}x}}\\{\coth ^2}x - 1 &= {{\mathop{\rm csch}\nolimits} ^2}x\end{aligned}\)

Thus, it is proved that \({\coth ^2}x - 1 = {{\mathop{\rm csch}\nolimits} ^2}x\).