Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

Answer: (B)

$2$

We know $\frac{\tan hx+1}{\tan hx-1}$ $=\frac{\frac{\sin hx}{\cos hx}+1}{\frac{\sin hx}{\cos hx}-1}$

$=\frac{\sin hx+\cos hx}{\sin hx-\cos hx}$

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

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

$=-e^{2x}$

$=-e^{px}$

Hence, $-e^{2x}=-e^{px}$

$e^{2x}=e^{px}$

$p=2$