Question

If $\frac{tanhx+1}{tanhx-1}=-e^{px}$, then p=

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is B $2$

$\frac{\tanh x+1}{\tanh x-1}=\frac{\frac{\sinh x}{\cosh x}+1}{\frac{\sinh x}{\cosh x}-1}$

$=\frac{\sinh x+\cosh x}{\sinh x-\cosh x}$

$=\frac{\frac{e^x-e^{-x}}{2}+\frac{e^x+e^{-x}}{2}}{\frac{e^x-e^{-x}}{2}-\frac{e^x+e^{-x}}{2}}$

$=\frac{e^x}{-e^{-x}}$

$=-e^{2x}$

by comparing it with $-e^{px}$

we get $p=2$