Question

Prove Equation 5 using (a) the method of Example 3 and (b) Exercise 22 with x replaced by y.

Short answer

The value of \({\tanh ^{ - 1}}x\) from both the methods is \(\frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\).

Step-by-step solution

Step 1:Simplify the problem using the trigonometric identity

Let \(y = {\tanh ^{ - 1}}x\), then:

\(\begin{aligned}x &= \tanh y\\ &= \frac{{\sinh y}}{{\cosh y}}\\ &= \frac{{\frac{{{e^y} - {e^{ - y}}}}{2}}}{{\frac{{{e^y} + {e^{ - y}}}}{2}}}\\ &= \frac{{{e^y} - {e^{ - y}}}}{{{e^y} + {e^{ - y}}}}\end{aligned}\)

Solve the equation in step 1 by solving exponents

Solve the equation \(x = \frac{{{e^y} - {e^{ - y}}}}{{{e^y} + {e^{ - y}}}}\).

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

Solve the equation in step 1 using Example 22

Using example 22,

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

So, the value of \({\tanh ^{ - 1}}x\) from both the methods is \(\frac{1}{2}\ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\).