Express the value of the inverse hyperbolic function in terms of natural logarithms.
-$\operatorname { coth } ^ { - 1 } \left( \frac { 6 } { 5 } \right)$
A) $\frac { 1 } { 2 } \ln 11$
B) $\frac { 1 } { 2 } \ln 66$
C) 0
D) $\frac { 1 } { 2 } \ln - 11$

Question

Quizplus · Accepted Answer

We know that $\operatorname{coth}^{-1}(x)=\frac{1}{2}\ln\left(\frac{x+1}{x-1}\right)$.
Using this formula, $\operatorname{coth}^{-1}\left(\frac{6}{5}\right)=\frac{1}{2}\ln\left(\frac{\frac{6}{5}+1}{\frac{6}{5}-1}\right)=\frac{1}{2}\ln(11)$. Therefore, the answer is A.

Express the Value of the Inverse Hyperbolic Function in Terms coth⁡−1(65)\operatorname { coth } ^ { - 1 } \left( \frac { 6 } { 5 } \right)coth−1(56​)

Express the Value of the Inverse Hyperbolic Function in Terms $\operatorname { coth } ^ { - 1 } \left( \frac { 6 } { 5 } \right)$