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Question: \(\frac{1 + {\tan h}x}{1 - {\tan h}x}\) is equal to....

1+tan⁡hx1−tan⁡hx\frac{1 + {\tan h}x}{1 - {\tan h}x} is equal to.

A

e2xe^{2x}

B

e−2xe^{- 2x}

C

i

D

– 1

Answer

e2xe^{2x}

Explanation

Solution

(1+tanh⁡x1−tanh⁡x)=(1+ex−e−xex+e−x1−ex−e−xex+e−x)=ex+e−x+ex−e−xex+e−x−ex+e−x\left( \frac{1 + \tanh x}{1 - \tanh x} \right) = \left( \frac{1 + \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}}{1 - \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}} \right) = \frac{e^{x} + e^{- x} + e^{x} - e^{- x}}{e^{x} + e^{- x} - e^{x} + e^{- x}}

=(2ex2e−x)=e2x= \left( \frac{2e^{x}}{2e^{- x}} \right) = e^{2x}.