Question:
If
\[
\operatorname{cosech}x=\frac{4}{5},
\]
then
\[
\cosh x=
\]
is:
If
\[
\operatorname{cosech}x=\frac{4}{5},
\]
then
\[
\cosh x=
\]
is:
Show Hint
Remember the standard hyperbolic identity:
\[
\cosh^2x-\sinh^2x=1
\]
which is analogous to
\[
\cos^2\theta+\sin^2\theta=1
\]
in trigonometry.
Updated On: Jun 24, 2026
- \(\sqrt{\dfrac{41}{21}}\)
- \(\sqrt{\dfrac{41}{19}}\)
- \(\sqrt{\dfrac{41}{25}}\)
- \(\sqrt{\dfrac{41}{16}}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use the definition of cosech.
Given, \[ \operatorname{cosech}x=\frac{4}{5} \] Since \[ \operatorname{cosech}x=\frac{1}{\sinh x}, \] we get \[ \sinh x=\frac{5}{4} \]
Step 2: Use the hyperbolic identity.
We know that \[ \cosh^2x-\sinh^2x=1 \] Substitute \[ \sinh x=\frac{5}{4} \] Then, \[ \cosh^2x-\left(\frac{5}{4}\right)^2=1 \] \[ \cosh^2x-\frac{25}{16}=1 \] \[ \cosh^2x=1+\frac{25}{16} \] \[ \cosh^2x=\frac{16+25}{16} \] \[ \cosh^2x=\frac{41}{16} \]
Step 3: Find \(\cosh x\).
Taking positive square root, \[ \cosh x=\sqrt{\frac{41}{16}} \] \[ \cosh x=\frac{\sqrt{41}}{4} \] Thus, \[ \cosh x=\sqrt{\frac{41}{16}} \]
Step 4: Final conclusion.
Hence, \[ \boxed{\sqrt{\dfrac{41}{16}}} \]
Given, \[ \operatorname{cosech}x=\frac{4}{5} \] Since \[ \operatorname{cosech}x=\frac{1}{\sinh x}, \] we get \[ \sinh x=\frac{5}{4} \]
Step 2: Use the hyperbolic identity.
We know that \[ \cosh^2x-\sinh^2x=1 \] Substitute \[ \sinh x=\frac{5}{4} \] Then, \[ \cosh^2x-\left(\frac{5}{4}\right)^2=1 \] \[ \cosh^2x-\frac{25}{16}=1 \] \[ \cosh^2x=1+\frac{25}{16} \] \[ \cosh^2x=\frac{16+25}{16} \] \[ \cosh^2x=\frac{41}{16} \]
Step 3: Find \(\cosh x\).
Taking positive square root, \[ \cosh x=\sqrt{\frac{41}{16}} \] \[ \cosh x=\frac{\sqrt{41}}{4} \] Thus, \[ \cosh x=\sqrt{\frac{41}{16}} \]
Step 4: Final conclusion.
Hence, \[ \boxed{\sqrt{\dfrac{41}{16}}} \]
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