Question

If \( 2\tanh^{-1}x = \sinh^{-1}\left(\frac{4}{3}\right) \) then \( \cosh^{-1}\left(\frac{1}{x}\right) = \)

• A. \( \log(\sqrt{2}+1) \)
• B. \( \log(\sqrt{2}-1) \)
• C. \( \log(2+\sqrt{3}) \)
• D. \( \log(2-\sqrt{3}) \)

Hint:

Remember the identity \( \tanh(\ln k) = \frac{k^2-1}{k^2+1} \). It directly converts log arguments to hyperbolic values.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

Convert the inverse hyperbolic functions to their logarithmic forms to solve for \( x \), then compute the required expression.
Step 2: Key Formula or Approach:

\( \sinh^{-1} z = \ln(z + \sqrt{z^2+1}) \)
\( \cosh^{-1} z = \ln(z + \sqrt{z^2-1}) \)
\( \tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \)
Step 3: Detailed Explanation:

Evaluate RHS: \[ \sinh^{-1}\left(\frac{4}{3}\right) = \ln\left(\frac{4}{3} + \sqrt{\frac{16}{9}+1}\right) = \ln\left(\frac{4}{3} + \frac{5}{3}\right) = \ln(3) \] Solve for x: \[ 2\tanh^{-1}x = \ln 3 \] \[ \tanh^{-1}x = \ln \sqrt{3} \] Let \( \tanh^{-1}x = y \implies x = \tanh y \). \[ x = \tanh(\ln \sqrt{3}) = \frac{e^{2\ln\sqrt{3}}-1}{e^{2\ln\sqrt{3}}+1} = \frac{3-1}{3+1} = \frac{2}{4} = \frac{1}{2} \] Evaluate Target Expression: We need \( \cosh^{-1}\left(\frac{1}{x}\right) = \cosh^{-1}(2) \). \[ \cosh^{-1}(2) = \ln(2 + \sqrt{2^2-1}) = \ln(2 + \sqrt{3}) \]
Step 4: Final Answer:

The value is \( \log(2+\sqrt{3}) \).