Q67 / 160 · AP EAMCET 2013 · Mathematics

$\tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth}^{-1}(2)$ is equal to

  • A
    $\frac{1}{2} \log 3$
  • B
    $\frac{1}{2} \log 6$
  • C
    $\frac{1}{2} \log 12$
  • Correct answer
    D
    $\log 3$

Solution

$\begin{aligned} & \tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth}^{-1}(2) \\ & \Rightarrow \tanh ^{-1}\left(\frac{1}{2}\right)+\tanh ^{-1}\left(\frac{1}{2}\right) \\ & \Rightarrow 2 \tanh ^{-1}\left(\frac{1}{2}\right) \quad\left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)\left[\tanh ^{-1} x=\frac{1}{2} \log \left(\frac{1+x}{1-x}\right)\right] \\ & \Rightarrow 2 \cdot \frac{1}{2} \log \left(\tanh ^{-1}(1 / x)\right] \\ & =\log \left(\frac{\frac{3}{2}}{\frac{2}{2}}\right)=\log 3\end{aligned}$

Hence, the correct answer is option: 'D'.

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