Question

The value of
\[ \frac{1+\tanh x}{1-\tanh x} \] is

• A. \(e^x\)
• B. \(e^{-2x}\)
• C. \(e^{2x}\)
• D. \(e^{-x}\)

Hint:

Remember the important identity: \(\displaystyle \frac{1+\tanh x}{1-\tanh x}=e^{2x}\). It is obtained directly from the exponential form of \(\tanh x\).

Answer 1

The Correct Option is C

Step 1: Use the definition of \(\tanh x\).
We know that
\[ \tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}} \]
Substitute this into the given expression:
\[ \frac{1+\tanh x}{1-\tanh x} = \frac{ 1+\frac{e^x-e^{-x}}{e^x+e^{-x}} }{ 1-\frac{e^x-e^{-x}}{e^x+e^{-x}} } \]

Step 2: Simplify numerator and denominator.
Numerator:
\[ 1+\frac{e^x-e^{-x}}{e^x+e^{-x}} = \frac{e^x+e^{-x}+e^x-e^{-x}}{e^x+e^{-x}} \]
\[ = \frac{2e^x}{e^x+e^{-x}} \]
Denominator:
\[ 1-\frac{e^x-e^{-x}}{e^x+e^{-x}} = \frac{e^x+e^{-x}-e^x+e^{-x}}{e^x+e^{-x}} \]
\[ = \frac{2e^{-x}}{e^x+e^{-x}} \]

Step 3: Divide the two fractions.
\[ \frac{\frac{2e^x}{e^x+e^{-x}}}{\frac{2e^{-x}}{e^x+e^{-x}}} \]
Cancelling common factors,
\[ =\frac{e^x}{e^{-x}} \]
\[ =e^{2x} \]

Step 4: Final conclusion.
Hence,
\[ \boxed{e^{2x}} \]