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