{2}+\operatorname{coth}^{-1} 3=$

{2}+\operatorname{coth}^{-1} 3=$
  1. $\log \sqrt{6}$
  2. $\log 6$
  3. $-\log \sqrt{6}$
  4. $-\log 6$

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)

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