Question

Evaluate: \(\tanh^{-1}\left(\frac{1}{3}\right) + \coth^{-1}(3) =\)

• A. \(\sech^{-1}\left(\frac{1}{3}\right)\)
• B. \(\cosech^{-1}\left(\frac{1}{3}\right)\)
• C. \(\cosh^{-1}\left(\frac{4}{3}\right)\)
• D. \(\sinh^{-1}\left(\frac{3}{4}\right)\)

Hint:

Remember the relationships between inverse hyperbolic functions and use the double argument formula to simplify sums.

Answer 1

The Correct Option is D

Recall that \(\coth^{-1}(x) = \tanh^{-1}\left(\frac{1}{x}\right)\) for \(x>1\). Thus, \[ \tanh^{-1}\left(\frac{1}{3}\right) + \coth^{-1}(3) = \tanh^{-1}\left(\frac{1}{3}\right) + \tanh^{-1}\left(\frac{1}{3}\right) = 2 \tanh^{-1}\left(\frac{1}{3}\right). \] Using the double argument formula for \(\tanh^{-1}\), \[ 2\tanh^{-1}(x) = \sinh^{-1}\left(\frac{2x}{1 - x^2}\right). \] Put \(x = \frac{1}{3}\), \[ \frac{2 \times \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2} = \frac{\frac{2}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4}. \] Hence, \[ \tanh^{-1}\left(\frac{1}{3}\right) + \coth^{-1}(3) = \sinh^{-1}\left(\frac{3}{4}\right). \]