Question

\(\coth^{-1}2=\)

• A. \(\operatorname{sech}^{-1}\!\left(\dfrac1{\sqrt5}\right)\)
• B. \(\tanh^{-1}\!\left(\dfrac13\right)\)
• C. \(\cosh^{-1}(\sqrt2)\)
• D. \(\operatorname{cosech}^{-1}(\sqrt3)\)

Hint:

Remember the identity \[ \boxed{\coth^2x-\operatorname{cosech}^2x=1.} \] It is the hyperbolic analogue of \[ \cot^2x-\csc^2x=-1. \]

Answer 1

The Correct Option is D

Step 1: Assume \[ \coth^{-1}2=t. \] Then, \[ \coth t=2. \]

Step 2: Use the hyperbolic identity. We know that \[ \coth^2t-\operatorname{cosech}^2t=1. \] Substituting \(\coth t=2\), \[ 4-\operatorname{cosech}^2t=1. \] Hence, \[ \operatorname{cosech}^2t=3. \] Therefore, \[ \operatorname{cosech}t=\sqrt3. \]

Step 3: Express \(t\) in inverse form. Thus, \[ t=\operatorname{cosech}^{-1}(\sqrt3). \] Since \[ t=\coth^{-1}2, \] we obtain \[ \boxed{\coth^{-1}2=\operatorname{cosech}^{-1}(\sqrt3).} \]