Question

The value of $2cot{h}^{-1}\left(\frac{z}{2}\right)$ is

• A. $\left[\ln \left(\frac{z-1}{z+2}\right)\right]$
• B. $\frac{1}{2}\left[\ln \left(\frac{z-1}{z+2}\right)\right]$
• C. $\frac{1}{2}\left[\ln \left(\frac{z+1}{z-1}\right)\right]$
• D. $-\left[\ln \left(\frac{z-2}{z+2}\right)\right]$

Answer 1

The correct option is D

$-\left[\ln \left(\frac{z-2}{z+2}\right)\right]$

Explanation for the correct answer:

The inverse hyperbolic cotangent function id\s defined as

$cot{h}^{-1}\left(z\right)=\frac{1}{2}\left[\ln \left(1+\frac{1}{z}\right)-\ln \left(1-\frac{1}{z}\right)\right]$

Substitute $\frac{z}{2}$

$cot{h}^{-1}\left(\frac{z}{2}\right)=\frac{1}{2}\left[\ln \left(1+\frac{2}{z}\right)-\ln \left(1-\frac{2}{z}\right)\right]$

$=\frac{1}{2}\left[\ln \left(\frac{z+2}{z}\right)-\ln \left(\frac{z-2}{z}\right)\right]$

$=\frac{1}{2}\left[\ln \left(\left(\frac{z+2}{z}\right)/\left(\frac{z-2}{z}\right)\right)\right]$

$\Rightarrow$ $cot{h}^{-1}\left(\frac{z}{2}\right)=\frac{1}{2}\left[\ln \left(\frac{z+2}{z-2}\right)\right]$

$\Rightarrow$$2cot{h}^{-1}\left(\frac{z}{2}\right)=\left[\ln \left(\frac{z+2}{z-2}\right)\right]$

$\Rightarrow$$2cot{h}^{-1}\left(\frac{z}{2}\right)=-\left[\ln \left(\frac{z-2}{z+2}\right)\right]$

Hence, option $\left(D\right)$ is the correct answer.