Q65 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics
$\operatorname{coth}^{-1} 3+\tanh ^{-1} \frac{1}{3}-\operatorname{cosech}^{-1}(-\sqrt{3})=$
- A$\log _e\left(\frac{2}{\sqrt{3}}\right)$
- Correct answerB$\log _e 2 \sqrt{3}$
- C0
- D$\log _8 3 \sqrt{3}$
Solution
We know that, $\operatorname{coth}^{-1} x=\frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right),|x|>1$
$\begin{aligned}
& \tanh ^{-1} x=\frac{1}{2} \log _e\left(\frac{1+x}{1-x}\right) \\
& \text {and } \operatorname{cosech}^{-1} x=\log _e\left(\frac{1-\sqrt{1+x^2}}{x}\right), x < 0 . \\
& \therefore \operatorname{coth}^{-1}(3)=\frac{1}{2} \log _e\left(\frac{4}{2}\right)=\log _e \sqrt{2} \\
& \tanh ^{-1}\left(\frac{1}{3}\right)=\frac{1}{2} \log _e\left(\frac{1+1 / 3}{1-1 / 3}\right)=\frac{1}{2} \log _e\left(\frac{4}{2}\right)=\log _e \sqrt{2} \\
& \text {and } \operatorname{cosech}^{-1}(-\sqrt{3})=\log _e\left(\frac{1-\sqrt{1+3}}{-\sqrt{3}}\right) \\
& =\log _e\left(\frac{1-2}{-\sqrt{3}}\right)=\log _e\left(\frac{1}{\sqrt{3}}\right)
\end{aligned}$
Hence, $\operatorname{coth}^{-1}(3)+\tanh ^{-1}\left(\frac{1}{3}\right)-\operatorname{cosech}^{-1}(-\sqrt{3})$
$\begin{aligned}
& =\log _e \sqrt{2}+\log _e \sqrt{2}-\log _e\left(\frac{1}{\sqrt{3}}\right) \\
& =\log _e(\sqrt{2} \cdot \sqrt{2})+\log _e(\sqrt{3})=\log _e(2 \sqrt{3})
\end{aligned}$
Hence, the correct answer is option: 'B'.