Question

(a) To prove the formula\({\tanh ^{ - 1}}x = \frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\)using the method of Example 3.

(b) To prove the formula\({\tanh ^{ - 1}}x = \frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\)by replacing\(x\)by\(y\)in Exercise 14.

Short answer

1. The proof is given below.
2. The proof is given below.

Step-by-step solution

Given data

The equation to be proved is \({\tanh ^{ - 1}}x = \frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\).

Formula of hyperbolic function

The identity is\(\frac{{1 + \tanh y}}{{1 - \tanh y}} = {e^{2y}}\).

Prove of the equation \({\tanh ^{ - 1}}x = \frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\)by the example 3 method

(a)

To prove the given equation in example 3 method let\(y = {\tanh ^{ - 1}}x\).

Then,\(x = \tanh y\).

Now, solve for\(y\).

\(\begin{array}{l}x = \tanh y\\x = \frac{{\sinh y}}{{\cosh y}}\\x = \frac{{{e^y} - {e^{ - y}}}}{{{e^y} + {e^{ - y}}}}\\x\left( {{e^y} + {e^{ - y}}} \right) = {e^y} - {e^{ - y}}\end{array}\)

Simplification of equation\(x\left( {{e^y} + {e^{ - y}}} \right) = {e^y} - {e^{ - y}}\)

Simplify further as shown below.

\(\begin{array}{c} - {e^y} + x{e^y} - {e^{ - y}} - x{e^{ - y}}\\{e^y}( - 1 + x) = - {e^{ - y}}(1 + x)\\{e^y}{e^{ - y}}(x - 1) = - {e^{ - y}}{e^{ - y}}(1 + x)\\(x - 1) = - {e^{ - 2y}}(1 + x)\end{array}\)

(Multiply both side by \({e^{ - y}}\))

Solve for\(y\).

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

So,\({\tanh ^{ - 1}}x = \frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\).

Therefore, the formula for \({\tanh ^{ - 1}}x\) is \({\tanh ^{ - 1}}x = \frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right).\)

Prove of the equation \({\tanh ^{ - 1}}x = \frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\) by exercise 14 method

(b)

To prove the given equation let\(y = {\tanh ^{ - 1}}x\).

Then,\(x = \tanh y\).

Substitute\(x = \tanh y\)in the identity\(\frac{{1 + \tanh y}}{{1 - \tanh y}} = {e^{2y}}\).

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

Therefore, the formula for \({\tanh ^{ - 1}}x\) is \({\tanh ^{ - 1}}x = \frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\).