Question

Prove the identity\(\frac{{{1 + tanhx}}}{{{1 - tanhx}}}{ = }{{e}^{{2x}}}\).

Short answer

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

Step-by-step solution

Given identity

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

The concept of the hyperbolic function

(1) The hyperbolic sine function:\({sinhx = }\frac{{{{e}^{x}}{ - }{{e}^{{ - x}}}}}{{2}}\)

(2) The hyperbolic cosine function:\({coshx = }\frac{{{{e}^{x}}{ + }{{e}^{{ - x}}}}}{{2}}\)

(3) The hyperbolic tangent function:\({tanhx = }\frac{{{sinhx}}}{{{coshx}}}\)

Use the formulas and prove

Consider, \(\frac{{1 + \tanh x}}{{1 - \tanh x}}\).

Use definition (3) and simplify.

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

Use the definition (1) and (2).

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

Hence, the required result \(\frac{{1 + \tanh x}}{{1 - \tanh x}} = {e^{2x}}\) is proved.