TS EAMCET · Maths · Inverse Trigonometric Functions
\(\operatorname{sech}^{-1}\left(\frac{3}{5}\right)-\tanh ^{-1}\left(\frac{3}{5}\right)=\)
- A \(\log_e 6\)
- B \(\log_e 5\)
- C \(\log _e\left(\frac{3}{2}\right)\)
- D \(\log _e\left(\frac{2}{3}\right)\)
Answer & Solution
Correct Answer
(C) \(\log _e\left(\frac{3}{2}\right)\)
Step-by-step Solution
Detailed explanation
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, \(\cosh ^{-1} x=\log \left(x+\sqrt{x^2-1}\right)\)…
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