Question

Prove the identity.

22. \(\frac{{{\bf{1}} + {\bf{tanh}}\,x}}{{{\bf{1}} - {\bf{tanh}}\,x}} = {e^{{\bf{2}}x}}\)

Short answer

The given identity is true.

Step-by-step solution

Step 1:Solve the expression \(\frac{{{\bf{1}} + {\bf{tanh}}\,x}}{{{\bf{1}} - {\bf{tanh}}\,x}}\)

Solve the expression \(\tanh \left( {\ln x} \right)\) using trigonometric relations.

\begin{aligned}\frac{{1 + \tanh x}}{{1 - \tanh x}} &= \frac{{1 + \frac{{\sinh x}}{{\cosh x}}}}{{1 - \frac{{\sinh x}}{{\cosh x}}}}\\ &= \frac{{\cosh x + \sinh x}}{{\cosh x - \sinh x}}\end{aligned}\)

Solve the equation in step 1 using the relation of hyperbolic functions

Simplify the equation \(\frac{{1 + \tanh x}}{{1 - \tanh x}} = \frac{{\cosh x + \sinh x}}{{\cosh x - \sinh x}}\) using properties of the logarithm.

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

Hence proved, the given identity is true.