{2}+\operatorname{coth}^{-1} 3=$
{2}+\operatorname{coth}^{-1} 3=$
- $\log \sqrt{6}$
- $\log 6$
- $-\log \sqrt{6}$
- $-\log 6$
Solution
$
\begin{aligned}
& \text {} \tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth} h^{-1} \text { (3) } \\
& =\frac{1}{2} \ln \left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)+\frac{1}{2} \ln \left(\frac{3+1}{3-1}\right)=\frac{1}{2} \log \left(\frac{\frac{3}{2}}{\frac{1}{2}}\right)+\frac{1}{2} \log \left(\frac{4}{2}\right) \\
& =\frac{1}{2} \log 3+\frac{1}{2} \log 2 \\
& =\log \sqrt{3}+\log \sqrt{2} \\
& =\log (\sqrt{3} \cdot \sqrt{2})=\log \sqrt{6} \text {. } \\
\end{aligned}
$
Asked in: AP EAMCET 2018 (22 Apr Shift 1)
Practice more Trigonometric Functions questions on Aicharya