Q67 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
$\operatorname{sech}^{-1}\left(\frac{3}{5}\right)-\tanh ^{-1}\left(\frac{3}{5}\right)=$
- A$\log_e 6$
- B$\log_e 5$
- Correct answerC$\log _e\left(\frac{3}{2}\right)$
- D$\log _e\left(\frac{2}{3}\right)$
Solution
To find : $\operatorname{sech}^{-1}\left(\frac{3}{5}\right)+\tanh ^{-1}\left(\frac{3}{5}\right)$
We know, $\operatorname{sech}^{-1}(x)=\cosh ^{-1}\left(\frac{1}{x}\right)$
Now, $\cosh ^{-1} x=\log \left(x+\sqrt{x^2-1}\right)$
$\begin{aligned}
& \cosh ^{-1} \frac{5}{3}=\log \left(\frac{5}{3}+\sqrt{\frac{25}{9}-1}\right)=\log \left(\frac{5}{3}+\frac{4}{9}\right)=\log 3 \\
& \therefore \operatorname{sech}^{-1}\left(\frac{3}{5}\right)=\cosh { }^{-1}\left(\frac{5}{3}\right)=\log 3
\end{aligned}$
Also, $\tan \mathrm{h}^{-1}(x)=\frac{1}{2} \log \left(\frac{1-x}{1+x}\right)$ $\tan \mathrm{h}^{-1}\left(\frac{3}{5}\right)=\frac{1}{2} \log \left(\frac{1-\frac{3}{5}}{1+\frac{3}{5}}\right)$ $=\frac{1}{2} \log \frac{2}{8}=\frac{1}{2} \log \frac{1}{4}=\log \frac{1}{2}$ Hence, $\operatorname{sech}^{-1}\left(\frac{3}{5}\right)+\tan h^{-1}\left(\frac{3}{5}\right)=\log 3+\log \frac{1}{2}=\log \frac{3}{2} .$
Also, $\tan \mathrm{h}^{-1}(x)=\frac{1}{2} \log \left(\frac{1-x}{1+x}\right)$ $\tan \mathrm{h}^{-1}\left(\frac{3}{5}\right)=\frac{1}{2} \log \left(\frac{1-\frac{3}{5}}{1+\frac{3}{5}}\right)$ $=\frac{1}{2} \log \frac{2}{8}=\frac{1}{2} \log \frac{1}{4}=\log \frac{1}{2}$ Hence, $\operatorname{sech}^{-1}\left(\frac{3}{5}\right)+\tan h^{-1}\left(\frac{3}{5}\right)=\log 3+\log \frac{1}{2}=\log \frac{3}{2} .$
Hence, the correct answer is option: 'C'.