Question

If \( x = \log_e 3 \), then \( \tanh(2x) + \operatorname{sech}(2x) = \)

• A. \( \frac{4}{3} \)
• B. \( \frac{49}{41} \)
• C. \( \frac{4}{5} \)
• D. \( \frac{41}{49} \)

Hint:

A useful identity for hyperbolic functions is \( \tanh(x) + \operatorname{sech}(x) = \frac{\sinh x + 1}{\cosh x} \). Also remember the identity \( \cosh^2(x) - \sinh^2(x) = 1 \), which is analogous to the Pythagorean identity in trigonometry.

Answer 1

The Correct Option is B

Given \( x = \ln 3 \), we can find the values of \( e^x \) and \( e^{-x} \).
\( e^x = e^{\ln 3} = 3 \).
\( e^{-x} = \frac{1}{e^x} = \frac{1}{3} \).
We need to evaluate \( \tanh(2x) + \operatorname{sech}(2x) \).
Let's find \( \cosh(2x) \) and \( \sinh(2x) \) first using their exponential definitions.
\( e^{2x} = (e^x)^2 = 3^2 = 9 \).
\( e^{-2x} = (e^{-x})^2 = (1/3)^2 = 1/9 \).
\( \cosh(2x) = \frac{e^{2x} + e^{-2x}}{2} = \frac{9 + 1/9}{2} = \frac{82/9}{2} = \frac{41}{9} \).
\( \sinh(2x) = \frac{e^{2x} - e^{-2x}}{2} = \frac{9 - 1/9}{2} = \frac{80/9}{2} = \frac{40}{9} \).
Now we can find \( \tanh(2x) \) and \( \operatorname{sech}(2x) \).
\( \tanh(2x) = \frac{\sinh(2x)}{\cosh(2x)} = \frac{40/9}{41/9} = \frac{40}{41} \).
\( \operatorname{sech}(2x) = \frac{1}{\cosh(2x)} = \frac{1}{41/9} = \frac{9}{41} \).
Finally, add the two values.
\( \tanh(2x) + \operatorname{sech}(2x) = \frac{40}{41} + \frac{9}{41} = \frac{49}{41} \).