Answer:
Option B
Explanation:
We know that , $\cot h^{-1}x= \frac{1}{2}\log _{e}\left(\frac{x+1}{x-1}\right),|x|>1$
$\tan h^{-1}x= \frac{1}{2}\log _{e}\left(\frac{1+x}{1-x}\right)$
and $cosec h^{-1}x=\log _{e}\left(\frac{1-\sqrt{1+x^{2}}}{x}\right),x<0$
$\therefore$ $\cot h^{-1}(3)=\frac{1}{2}\log _{e}\left(\frac{4}{2}\right)=\log_{e} \sqrt{2}$
$\tan h^{-1}\left(\frac{1}{3}\right)=\frac{1}{2}\log _{e}\left(\frac{1+1/3}{1-1/3}\right)=\frac{1}{2}\log_{e}\left(\frac{4}{2}\right)=\log_{e} \sqrt{2}$
and $cosec h^{-1}(-\sqrt{3})=\log_{e}\left(\frac{1-\sqrt{1+3}}{-\sqrt{3}}\right)$
= $\log_{e}\left(\frac{1-2}{-\sqrt{3}}\right)=\log_{e}\left(\frac{1}{\sqrt{3}}\right)$
Hence, $\cot h^{-1}(3)+ \tanh^{-1} \frac{1}{3}- cosec h^{-1} (-\sqrt{3})=$
= $\log _{e}\sqrt{2}+\log_{e}\sqrt{2}-\log_{e}\left(\frac{1}{\sqrt{3}}\right)$
$=\log _{e}(\sqrt{2}.\sqrt{2})+\log_{e}(\sqrt{3})=\log_{e}(2\sqrt{3})$
