Chapter 5.2, Problem 80E

Solution

a)

To prove: The identity ${\cosh }^{2}x-{\sinh }^{2}x=1$ .

Given information:

The given identity is ${\cosh }^{2}x-{\sinh }^{2}x=1$ .

The hyperbolic trigonometric functions are $\sinh x=\frac{e^x-e^{-x}}{2}$ and $\cosh x=\frac{e^x+e^{-x}}{2}$

Formula used:

The hyperbolic trigonometric identities , $\sinh x=\frac{e^x-e^{-x}}{2}$ and $\cosh x=\frac{e^x+e^{-x}}{2}$ .

The algebraic identities ${\left(a+b\right)}^2=a^2+b^2+2ab$ and ${\left(a-b\right)}^2=a^2+b^2-2ab$ .

The exponential rules are $e^x{e}^y=e^{x+y}$ and $e^0=1$ .

Proof:

Simplify the left-hand side of the given equation using the hyperbolic trigonometric functions and algebraic identities as follows.

  $\begin{array}{c}{\cosh }^{2}x-{\sinh }^{2}x={\left(\frac{e^x+e^{-x}}{2}\right)}^2-{\left(\frac{e^x-e^{-x}}{2}\right)}^2\\ =\left(\frac{e^{2x}+e^{-2x}+2e^x{e}^{-x}}{4}\right)-\left(\frac{e^{2x}+e^{-2x}-2e^x{e}^{-x}}{4}\right)\\ =\frac{e^{2x}+e^{-2x}+2e^x{e}^{-x}-e^{2x}-e^{-2x}+2e^x{e}^{-x}}{4}\\ =\frac{4e^x{e}^{-x}}{4}\end{array}$

Again simplify the obtained expression using the exponential properties as follows.

  $\begin{array}{c}{\cosh }^{2}x-{\sinh }^{2}x=e^x{e}^{-x}\\ =e^{x-x}\\ =e^0\\ =1\end{array}$

Here, the left-hand side is equal to the right-hand side.

Hence, it is proved that ${\cosh }^{2}x-{\sinh }^{2}x=1$ .

b)

To prove: The identity $1-{\tanh }^{2}x={\mathrm{sech}}^{2}x$ .

Given information:

The given identity is $1-{\tanh }^{2}x={\mathrm{sech}}^{2}x$ .

The hyperbolic trigonometric functions are $\sinh x=\frac{e^x-e^{-x}}{2}$, $\cosh x=\frac{e^x+e^{-x}}{2}$, $\mathrm{sech}x=\frac{1}{\cosh x}$, and $\tanh x=\frac{\sinh x}{\cosh x}$ .

Formula used:

The hyperbolic trigonometric identities, $\tanh x=\frac{\sinh x}{\cosh x}$, $\mathrm{sech}x=\frac{1}{\cosh x}$, $\sinh x=\frac{e^x-e^{-x}}{2}$ and $\cosh x=\frac{e^x+e^{-x}}{2}$ .

The algebraic identities ${\left(a+b\right)}^2=a^2+b^2+2ab$ and ${\left(a-b\right)}^2=a^2+b^2-2ab$ .

The exponential rules are $e^x{e}^{-x}=1$ and $e^0=1$ .

Proof:

Simplify the left-hand side of the given equation using the hyperbolic trigonometric functions and algebraic identities as follows.

  $\begin{array}{c}1-{\tanh }^{2}x=1-{\left(\frac{\sinh x}{\cosh x}\right)}^2\\ =1-{\left(\frac{\frac{e^x-e^{-x}}{2}}{\frac{e^x+e^{-x}}{2}}\right)}^2\\ =1-{\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)}^2\\ =\frac{{\left(e^x+e^{-x}\right)}^2-{\left(e^x-e^{-x}\right)}^2}{{\left(e^x+e^{-x}\right)}^2}\end{array}$

Again simplify the obtained expression using the exponential properties as follows.

  $\begin{array}{c}1-{\tanh }^{2}x=\frac{e^{2x}+e^{-2x}+2e^x{e}^{-x}-e^{2x}-e^{-2x}+2e^x{e}^{-x}}{{\left(e^x+e^{-x}\right)}^2}\\ =\frac{4e^x{e}^{-x}}{{\left(e^x+e^{-x}\right)}^2}\\ =\frac{4}{{\left(e^x+e^{-x}\right)}^2}\\ ={\left(\frac{2}{e^x+e^{-x}}\right)}^2\end{array}$

Simplify the obtained expression using definition of hyperbolic trigonometric function. As follows.

  $\begin{array}{c}1-{\tanh }^{2}x={\left(\frac{1}{\cosh x}\right)}^2\\ ={\mathrm{sech}}^{2}x\end{array}$

Here, the left-hand side is equal to the right-hand side.

Hence, it is proved that $1-{\tanh }^{2}x={\mathrm{sech}}^{2}x$ .

c)

To prove: The identity ${\coth }^{2}x-1={\mathrm{csch}}^{2}x$ .

Given information:

The given identity is ${\coth }^{2}x-1={\mathrm{csch}}^{2}x$ .

The hyperbolic trigonometric functions are $\sinh x=\frac{e^x-e^{-x}}{2}$, $\cosh x=\frac{e^x+e^{-x}}{2}$, $\mathrm{sech}x=\frac{1}{\cosh x}$, and $\tanh x=\frac{\sinh x}{\cosh x}$ .

Formula used:

The hyperbolic trigonometric identities, $\tanh x=\frac{\sinh x}{\cosh x}$, $\mathrm{sech}x=\frac{1}{\cosh x}$, $\sinh x=\frac{e^x-e^{-x}}{2}$ and $\cosh x=\frac{e^x+e^{-x}}{2}$ .

The algebraic identities ${\left(a+b\right)}^2=a^2+b^2+2ab$ and ${\left(a-b\right)}^2=a^2+b^2-2ab$ .

The exponential rules are $e^x{e}^{-x}=1$ and $e^0=1$ .

Proof:

Simplify the left-hand side of the given equation using the hyperbolic trigonometric functions and algebraic identities as follows.

  $\begin{array}{c}{\coth }^{2}x-1={\left(\frac{\cosh x}{\sinh x}\right)}^2-1\\ ={\left(\frac{\frac{e^x+e^{-x}}{2}}{\frac{e^x-e^{-x}}{2}}\right)}^2-1\\ ={\left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right)}^2-1\\ =\frac{{\left(e^x+e^{-x}\right)}^2-{\left(e^x-e^{-x}\right)}^2}{{\left(e^x-e^{-x}\right)}^2}\end{array}$

Again simplify the obtained expression using the exponential properties as follows.

  $\begin{array}{c}{\coth }^{2}x-1=\frac{e^{2x}+e^{-2x}+2e^x{e}^{-x}-e^{2x}-e^{-2x}+2e^x{e}^{-x}}{{\left(e^x-e^{-x}\right)}^2}\\ =\frac{4e^x{e}^{-x}}{{\left(e^x-e^{-x}\right)}^2}\\ =\frac{4}{{\left(e^x-e^{-x}\right)}^2}\\ ={\left(\frac{2}{e^x-e^{-x}}\right)}^2\end{array}$

Simplify the obtained expression using definition of hyperbolic trigonometric function. As follows.

  $\begin{array}{c}{\coth }^{2}x-1={\left(\frac{1}{\sinh x}\right)}^2\\ ={\mathrm{csch}}^{2}x\end{array}$

Here, the left-hand side is equal to the right-hand side.

Hence, it is proved that ${\coth }^{2}x-1={\mathrm{csch}}^{2}x$ .