Question

Let \( x \in \mathbb{R} \) and \( |x|<1 \). Then \( \tanh^{-1} x = \) ?

• A. \( \frac{1}{2} \log\left( \frac{1 + x}{1 - x} \right) \)
• B. \( \frac{1}{2} \log\left( \frac{1 - x}{1 + x} \right) \)
• C. \( \frac{1}{2} \log\left( x + \sqrt{1 - x^2} \right) \)
• D. \( \frac{1}{2} \log\left( x - \sqrt{1 - x^2} \right) \)

Hint:

Remember the inverse identity: \( \tanh^{-1} x = \frac{1}{2} \log\left( \frac{1 + x}{1 - x} \right) \), valid only for \( |x|<1 \).

Answer 1

The Correct Option is A

The inverse hyperbolic tangent identity is: \[ \tanh^{-1} x = \frac{1}{2} \log\left( \frac{1 + x}{1 - x} \right), \quad \text{for } |x|<1 \] This is a standard formula from the definitions of inverse hyperbolic functions.