Question

If \(\sinh x=\frac{12}{5}\), then \[ \cosh2x-\sinh2x= \]

• A. 25
• B. \(\frac{1}{125}\)
• C. \(\frac{1}{25}\)
• D. 125 \bigskip

Hint:

Convert hyperbolic functions into exponentials for direct solving.

Answer 1

The Correct Option is C

Concept: Use identity: \[ \cosh x-\sinh x=e^{-x} \] Thus: \[ \cosh2x-\sinh2x=e^{-2x} \]

Step 1: Find \(e^x\).
\[ \sinh x=\frac{e^x-e^{-x}}{2}=\frac{12}{5} \] Let \(t=e^x\): \[ t-\frac1t=\frac{24}{5} \] \[ t^2-\frac{24}{5}t-1=0 \] \[ 5t^2-24t-5=0 \] \[ t=\frac{24\pm\sqrt{576+100}}{10} =\frac{24\pm26}{10} \] \[ t=5 \;(\text{positive}) \]

Step 2: Compute required value.
\[ e^{-2x}=\frac{1}{e^{2x}}=\frac{1}{25} \] \[ \boxed{(C)} \]