Chapter 2: Q. 99 (page 236)

Prove that the inverse hyperbolic functions can be written in terms of logarithms as shown in Exercises 97–99.
$\tan h^{- 1} x = \frac{1}{2} \ln (\frac{1 + x}{1 - x})$ , for $- 1 < x < 1$ .

Short Answer

Expert verified

We proved $\tan h^{- 1} x = \frac{1}{2} \ln (\frac{1 + x}{1 - x})$ , for $- 1 < x < 1$ .

Step by step solution

Step 1. Given Information

We have $\tan h y = x$ and we need to prove $t an h^{- 1} x = \frac{1}{2} \ln (\frac{1 + x}{1 - x})$ .

Step 2. Proving the statement

$\tan h y = x \Rightarrow \frac{\sin h y}{\cos h y} = x \Rightarrow \frac{\frac{e^{y} - e^{- y}}{2}}{\frac{e^{y} + e^{- y}}{2}} = x \Rightarrow \frac{e^{y} - e^{- y}}{e^{y} + e^{- y}} = x \Rightarrow \frac{e^{y} - \frac{1}{e^{y}}}{e^{y} + \frac{1}{e^{y}}} = x \Rightarrow \frac{e^{2 y} - 1}{e^{2 y} + 1} = x \Rightarrow e^{2 y} - 1 = x (e^{2 y} + 1) \Rightarrow e^{2 y} - 1 = x e^{2 y} + x \Rightarrow x e^{2 y} - e^{2 y} = - (x + 1) \Rightarrow e^{2 y} (x - 1) = - (x + 1) \Rightarrow e^{2 y} = \frac{- (x + 1)}{(x - 1)} \Rightarrow e^{2 y} = \frac{(x + 1)}{(1 - x)} \Rightarrow \ln (e^{2 y}) = \ln (\frac{(x + 1)}{(1 - x)}) \Rightarrow 2 y = \ln (\frac{(x + 1)}{(1 - x)}) \Rightarrow y = \frac{1}{2} \ln (\frac{(x + 1)}{(1 - x)}) \Rightarrow \tan h^{- 1} x = \frac{1}{2} \ln (\frac{(x + 1)}{(1 - x)})$

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Prove that the inverse hyperbolic functions can be written in terms of logarithms as shown in Exercises 97–99.
$\tan h^{- 1} x = \frac{1}{2} \ln (\frac{1 + x}{1 - x})$ , for $- 1 < x < 1$ .

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Proving the statement

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Prove that the inverse hyperbolic functions can be written in terms of logarithms as shown in Exercises 97–99.tanh-1x=12ln1+x1-x, for -1<x<1.

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Proving the statement

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

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Study anywhere. Anytime. Across all devices.

Prove that the inverse hyperbolic functions can be written in terms of logarithms as shown in Exercises 97–99.
$\tan h^{- 1} x = \frac{1}{2} \ln (\frac{1 + x}{1 - x})$ , for $- 1 < x < 1$ .