Question

If \[ \tanh^{-1}(x)=\log\sqrt3 \] and \[ \cosh^{-1}(y)=\log(1+\sqrt2), \] then \[ \operatorname{sech}^{-1}(xy)= \]

• A. \(\log(1+\sqrt2)\)
• B. \(\log(\sqrt3+\sqrt6)\)
• C. \(\log\sqrt2\)
• D. \(\log\sqrt3\)

Hint:

Remember the standard result: \[ \cosh^{-1}(\sqrt2)=\log(1+\sqrt2). \] It appears frequently in inverse hyperbolic function problems.

Answer 1

The Correct Option is A

Concept: We use the definitions: \[ \tanh^{-1}x = \frac12 \log\left(\frac{1+x}{1-x}\right), \] and \[ \cosh^{-1}y = \log\left(y+\sqrt{y^2-1}\right). \] After finding \(x\) and \(y\), we compute \(xy\) and then determine \(\operatorname{sech}^{-1}(xy)\).

Step 1: Determine \(x\). Given, \[ \tanh^{-1}(x)=\log\sqrt3. \] Using the definition, \[ \frac12 \log\left(\frac{1+x}{1-x}\right) = \log\sqrt3. \] Multiplying by \(2\), \[ \log\left(\frac{1+x}{1-x}\right) = \log3. \] Hence, \[ \frac{1+x}{1-x}=3. \] \[ 1+x=3-3x. \] \[ 4x=2. \] \[ x=\frac12. \]

Step 2: Determine \(y\). Given, \[ \cosh^{-1}(y)=\log(1+\sqrt2). \] Using the well-known identity \[ \cosh^{-1}(\sqrt2) = \log(1+\sqrt2), \] we obtain \[ y=\sqrt2. \]

Step 3: Calculate \(xy\). \[ xy = \frac12\cdot\sqrt2 = \frac1{\sqrt2}. \]

Step 4: Evaluate \(\operatorname{sech}^{-1}(xy)\). Since \[ xy=\frac1{\sqrt2}, \] we need \[ \operatorname{sech}^{-1}\left(\frac1{\sqrt2}\right). \] Let \[ u=\operatorname{sech}^{-1}\left(\frac1{\sqrt2}\right). \] Then \[ \operatorname{sech}u=\frac1{\sqrt2} \] which implies \[ \cosh u=\sqrt2. \] Therefore, \[ u=\cosh^{-1}(\sqrt2). \] Using the standard value, \[ u=\log(1+\sqrt2). \] Conclusion: \[ \boxed{\operatorname{sech}^{-1}(xy)=\log(1+\sqrt2)} \]