Explain Inverse Hyperbolic Functions Formula (original) (raw)

Last Updated : 23 Jul, 2025

In mathematics, the inverse functions of hyperbolic functions are referred to as inverse hyperbolic functions or area hyperbolic functions. There are six inverse hyperbolic functions, namely, inverse hyperbolic sine, inverse hyperbolic cosine, inverse hyperbolic tangent, inverse hyperbolic cosecant, inverse hyperbolic secant, and inverse hyperbolic cotangent functions. These functions are depicted as sinh-1 x, cosh-1 x, tanh-1 x, csch-1 x, sech-1 x, and coth-1 x. With the help of an inverse hyperbolic function, we can find the hyperbolic angle of the corresponding hyperbolic function.

Table of Content

What are Inverse Hyperbolic Functions?

Inverse hyperbolic functions are the inverse functions of the hyperbolic functions: hyperbolic sine (sinh), hyperbolic cosine (cosh), hyperbolic tangent (tanh), hyperbolic cotangent (coth), hyperbolic secant (sech), and hyperbolic cosecant (csch).

These functions are important in various fields, including engineering, physics, and mathematics.

Inverse Hyperbolic Functions Table

Function Name Function Formula Domain Range
Inverse Hyperbolic Sine sinh-1 x ln[x + √(x2 + 1)] (-∞, ∞) (-∞, ∞)
Inverse Hyperbolic Cosine cosh-1x ln[x + √(x2 - 1) [1, ∞) [0, ∞)
Inverse hyperbolic tangent tanh-1 x ½ ln[(1 + x)/(1 - x)] (-1,1) (-∞, ∞)
Inverse hyperbolic cosecant csch-1 x ln[(1 + √(x2 + 1)/x] (-∞, ∞) (-∞, ∞)
Inverse hyperbolic secant sech-1 x ln[(1 + √(1 - x2)/x] (0, 1] [0, ∞)
Inverse hyperbolic cotangent coth-1 x ½ ln[(x + 1)/(x - 1)] (-∞, -1) or (1, ∞) (-∞, ∞)

Let's learn about them in detail.

**Inverse Hyperbolic Sine Function

The formula for inverse hyperbolic sine function is added below:

**sinh -1 x = ln[x + √(x 2 + 1)]

The proof for the same is,

**Proof:

Let sinh-1 x = z, where z ∈ R

⇒ x = sinh z

Using the sine hyperbolic function we get,

⇒ x = (ez - e-z)/2

⇒ 2x = ez - e-z

⇒ e2z - 2xez - 1 = 0

We know that roots of an equation ax2 + bx + c = 0 are x = [-b ± √(b2 - 4ac)]/2a

So, ez = x ± √(x2 + 1)

Since z is a real number, e must be a positive number.

Hence, ez = x + √(x2 + 1)

⇒ z = ln[x + √(x2 + 1)]

⇒ sinh-1 x = ln[x + √(x2 + 1)]

**sinh -1 x = ln[x + √(x 2 + 1)]

**Inverse Hyperbolic Cosine Function

The formula for inverse hyperbolic cosine function is added below:

**cosh -1 x = ln[x + √(x 2 - 1)]

The proof for the same is,

**Proof:

Let cosh-1 x = z, where z ∈ R

⇒ x = cosh z

Using the cosine hyperbolic function we get,

⇒ x = (ez + e-z)/2

⇒ 2x = ez + e-z

⇒ e2z - 2xez + 1 = 0

We know that roots of an equation ax2 + bx + c = 0 are x = [-b ± √(b2 - 4ac)]/2a

So, ez = x ± √(x2 - 1)

Since z is a real number, e must be a positive number.

Hence, ez = x + √(x2 - 1)

⇒ z = ln[x + √(x2 - 1)]

⇒ cosh-1 x = ln[x + √(x2 - 1)]

**cosh -1 x = ln[x + √(x 2 - 1)]

**Inverse Hyperbolic Tangent Function

The formula for inverse hyperbolic tangent function is added below:

**tanh -1 x = ½ ln[(1 + x)/(1 - x)] = ½ [ln(1 + x) - ln(1 - x)]

The proof for the same is,

**Proof:

Let tanh-1 x = z, where z ∈ R

⇒ x = tanh z

Using the tangent hyperbolic function we get,

tanh z = (ez - e-z)/(ez + e-z)

x = \left[\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}}\right]\times\left[\frac{e^{z}}{e^{z}}\right]

⇒ x = (e2z - 1)/(e2z + 1)

⇒ x (e2z + 1) = (e2z - 1)

⇒ (x - 1) e2z + (x + 1) = 0

⇒ e2z = -[(x +1)/(x - 1)]

⇒ e2z = [(x + 1)/(1 - x)]

⇒ 2z = ln [(x + 1)/(1 - x)]

⇒ z = ½ ln[(1 + x)/(1 - x)] = ½ [ln(1 + x) - ln(1 - x)]

⇒ tanh-1 x = ½ ln[(1 + x)/(1 - x)] = ½ [ln(1 + x) - ln(1 - x)]

**tanh -1 x = ½ ln[\frac{1+x}{1-x}**] = ½ [ln(1 + x) - ln(1 - x)]

**Inverse Hyperbolic Cosecant Function

The formula for inverse hyperbolic cosecant function is added below:

**csch -1 x = ln[(1 + √(x 2 + 1)/x]

The proof for the same is,

**Proof:

Let csch-1 x = z, where z ∈ R

⇒ x = csch z

Using the cosecant hyperbolic function we get,

csch z = 2/(ez - e-z)

⇒ x = 2/(ez - e-z)

⇒ x = \left[\frac{2}{e^{z}-e^{-z}}\right]\times\left[\frac{e^{z}}{e^{z}}\right]

⇒ x = 2ez/(e2z - 1)

⇒ x (e2z - 1) = 2ez

⇒ xe2z − 2ez - x = 0

We know that roots of an equation ax2 + bx + c = 0 are x = [-b ± √(b2 - 4ac)]/2a

⇒ ez = (1 + √(x2 + 1)/x

⇒ z = ln[ \frac{1+\sqrt{(1}+x^{2})}{x} ]

⇒ csch-1x = ln[ \frac{1+\sqrt{(1}+x^{2})}{x} ] = ln[1 + √(1 + x2)] - ln(x)

**csch -1 x = ln[\frac{1+\sqrt{(1}+x^{2})}{x} **] = ln[1 + √(1 + x 2 )] - ln(x)

**Inverse Hyperbolic Secant Function

The formula for inverse hyperbolic secant function is added below:

**sech -1 x = ln[(1 + √(1 - x 2 )/x]

The proof for the same is,

**Proof:

Let sech-1 x = z, where z ∈ R

⇒ x = sech z

Using the secant hyperbolic function we get,

sech z = 2/(ez + e-z)

⇒ x = 2/(ez + e-z)

⇒ x = \left[\frac{2}{e^{z}+e^{-z}}\right]\times\left[\frac{e^{z}}{e^{z}}\right]

⇒ x = 2ez/(e2z + 1)

⇒ x (e2z +1) = 2ez

⇒ xe2z − 2ez + x = 0

We know that roots of an equation ax2 + bx + c = 0 are x = [-b ± √(b2 - 4ac)]/2a

So, by simplifying we get,

ez =\frac{1+\sqrt{(1}-x^{2})}{x}

z = ln[ \frac{1+\sqrt{(1}-x^{2})}{x} ] = ln[1 + √(1 - x2)] - ln(x)

⇒ sech-1 x = ln[ \frac{1+\sqrt{(1}-x^{2})}{x} ] = ln[1 + √(1 - x2)] - ln(x)

**sech -1 x = ln[\frac{1+\sqrt{(1}-x^{2})}{x} **] = ln[1 + √(1 - x 2 )] - ln(x)

**Inverse Hyperbolic Cotangent Function

The formula for inverse hyperbolic cotangent function is added below:

**coth -1 x = ½ ln[(x + 1)/(x - 1)]

The proof for the same is,

**Proof:

Let coth-1 x = z, where z ∈ R

⇒ x = coth z

Using the cotangent hyperbolic function we get,

coth z = (ez + e-z)/(ez - e-z)

⇒ x = (ez + e-z)/(ez - e-z)

⇒ x = \left[\frac{e^{z}+e^{-z}}{e^{z}-e^{-z}}\right]\times\left[\frac{e^{z}}{e^{z}}\right]

⇒ x = (e2z + 1)/(e2z - 1)

⇒ x (e2z - 1) = (e2z + 1)

⇒ (x - 1) e2z - (x + 1) = 0

⇒ e2z = [(x +1)/(x - 1)]

⇒ 2z = ln [(x + 1)/(x - 1)]

⇒ z = ½ ln[(x + 1)/(x - 1)] = ½[ln(x + 1) - ln(x - 1)]

**coth -1 x = ½ ln[(x + 1)/(x - 1)] = ½ [ln(x + 1) - ln(x - 1)]

**Solved Examples on Inverse Hyperbolic Functions

**Inverse Hyperbolic Sine (arcsinh or sinh⁻¹)

**Example 1: Find sinh⁻¹(3)

**Solution:

**sinh⁻¹(3) = ln(3 + √(3² + 1))

**= ln(3 + √10)

**≈ 1.8184

**Example 2: Solve the equation: sinh(x) = 2

**Solution:

**x = sinh⁻¹(2)

**= ln(2 + √(2² + 1))

**= ln(2 + √5)

**≈ 1.4436

**Inverse Hyperbolic Cosine (arccosh or cosh⁻¹)

**Example 3: Find cosh⁻¹(4)

**Solution:

**cosh⁻¹(4) = ln(4 + √(4² - 1))

**= ln(4 + √15)

**≈ 2.0634

**Example 4: Solve the equation: cosh(x) = 3

**Solution:

**x = cosh⁻¹(3)

**= ln(3 + √(3² - 1))

**= ln(3 + 2√2)

**≈ 1.7627

**Inverse Hyperbolic Tangent (arctanh or tanh⁻¹)

**Example 5: Find tanh⁻¹(0.5)

**Solution:

**tanh⁻¹(0.5) = 1/2 * ln((1 + 0.5) / (1 - 0.5))

**= 1/2 * ln(3)

**≈ 0.5493

**Example 6: Solve the equation: tanh(x) = -0.8

**Solution:

**x = tanh⁻¹(-0.8)

**= 1/2 * ln((1 - 0.8) / (1 + 0.8))

**= 1/2 * ln(0.1111...)

**≈ -1.0986

**Example 7: Evaluate sinh⁻¹(sinh(π/4))

**Solution:

**sinh⁻¹(sinh(π/4)) = π/4

**This is because sinh⁻¹ and sinh are inverse functions, so they cancel each other out.

**Example 8: Solve the equation: 2cosh⁻¹(x) = 5

**Solution:

**2cosh⁻¹(x) = 5

**cosh⁻¹(x) = 5/2

**Now, we apply cosh to both sides:

**cosh(cosh⁻¹(x)) = cosh(5/2)

**x = cosh(5/2)

**We can calculate this:

**x = (e^(5/2) + e^(-5/2)) / 2 ≈ 6.1322

**Example 9: Find the value of tanh⁻¹(3/5) + tanh⁻¹(4/5)

**Solution:

**Let a = tanh⁻¹(3/5) and b = tanh⁻¹(4/5)

**Using the formula for tanh⁻¹ sum:

**tanh⁻¹(A) + tanh⁻¹(B) = tanh⁻¹((A+B)/(1+AB))

**Here, A = 3/5 and B = 4/5

**tanh⁻¹(3/5) + tanh⁻¹(4/5) = tanh⁻¹((3/5 + 4/5) / (1 + (3/5)(4/5)))

**= tanh⁻¹((7/5) / (1 + 12/25))

**= tanh⁻¹((7/5) / (37/25))

**= tanh⁻¹(35/37)

**≈ 1.4722

**Example 10: Simplify: arcsinh(e x - e -x )

**Solution:

**First, recall that sinh(x) = (e x - e -x ) / 2

**So, arcsinh(e x - e -x ) = arcsinh(2sinh(x))

**Now, we can use the identity arcsinh(2sinh(x)) = ln(e x + √(e 2x + 1))

**Therefore:

**arcsinh(e x - e -x ) = ln(e^x + √(e 2x + 1))

Practice Questions on Explain Inverse Hyperbolic Functions Formula

**1. Evaluate sinh⁻¹(0).

**2. Solve the equation: cosh(x) = 2.

**3. Find the value of tanh⁻¹(1/3).

**4. Simplify: cosh⁻¹(cosh(2x)).

**5. Solve the equation: sinh⁻¹(x) + sinh⁻¹(2x) = sinh⁻¹(5).

**6. Prove that tanh⁻¹(x) = 1/2 ln((1+x)/(1-x)) for |x| < 1.

**7. Find the derivative of f(x) = arcsinh(x).

**8. Evaluate the definite integral: ∫[0 to 1] (1 / √(1+x²)) dx.

**9. Solve the equation: e x + e -x = 3.

**10. Prove the identity: sinh⁻¹(x) = ln(x + √(x² + 1)).

**11. Evaluate the definite integral: ∫[0 to 1] (x / √(1+x²)) dx.

**12. Prove the identity: tanh⁻¹(x) = arcsinh(x/√(1+x²)).

**13. Find the derivative of f(x) = arccosh(x).

**14. Solve the equation: cosh⁻¹(x) + cosh⁻¹(2x) = cosh⁻¹(3).

**15. Prove the identity: tanh⁻¹(x) = arcsinh(x/√(x² + 1)).

**16. Solve the equation: e^(2x) + e^(-2x) = 7.

**17. Find the value of csch⁻¹(1/4).

**18.Prove that coth⁻¹(x) = 1/2 ln((x + 1)/(x - 1)) for |x| > 1.

**19. Simplify: sinh⁻¹(sinh(x) + sinh(y)).

**20. Solve the equation: tanh(x) = 2/3.

**Derivates of Inverse Hyperbolic Functions

Derivative of Inverse Hyperbolic Function is added in the table below:

Inverse Hyperbolic Function Derivative
sinh-1x 1/√(x2 + 1)
cosh-1 x 1/√(x2 - 1), x>1
tanh-1x 1/(1 - x2), |x
csch-1 x 1/{|x
sech-1 x -1/[x√(1 - x2)], 0 < x < 1
coth-1 x 1/(1 - x2), |x

**Read More:

Examples on Inverse Hyperbolic Functions Formula

**Example 1: If sinh x = 4, then prove that x = log e (4 + √17).

**Solution:

Given, sinh x = 4

⇒ x = sinh-1 (4)

We know that,

**sinh -1 (x) = loge [x + √(x 2 + 1)]

⇒ x = loge[4 + √(42 + 1)] = loge(4 + √17)

Hence, x = loge(4 + √17)

**Example 2: Prove that tanh -1 (sin x) = cosh -1 (sec x).

**Solution:

We know that,

**tanh -1 x = 1/2 ln[(1+x)/(1-x)]

Now, tanh-1 (sin x) = 1/2 log[(1 + sin x)/(1 - sin x)]

We have,

**cosh -1 x = ln(x + √[x 2 -1])

Now, cosh-1 (sec x) = ln[sec x + √(sec2 x - 1)]

= ln[sec x + √tan2 x] ****{Since, sec** 2 x - 1 = tan 2 x}

= ln[sec x + tan x]

= ln[1/cos x + sin x/cos x]

= ln[(1 + sin x)/cos x]

Now, multiply and divide the term with 2

= 1/2 × 2 ln[(1 + sin x)/cos x]

= 1/2 ln[(1 + sin x)/cos x]2 ****{since, 2 ln x = ln x** 2 }

= 1/2 ln[(1 + sin x)2/cos2 x]

We know, **cos 2 x = 1 - sin 2 x = (1 + sin x)(1 - sin x)

Hence, (1 + sin x)2/cos2 x = [(1 + sin x)(1+ sin x)]/[(1 + sin x)(1 - sin x)] = (1 + sin x)/(1 - sin x)

= 1/2 ln[(1 + sin x)/(1 - sin x)]

= tanh-1 (sin x)

Hence, tanh-1 (sin x) = cosh-1 (sec x)

**Example 3: Find the value of tanh -1 ****(1/5).**

**Solution:

We know,

**tanh -1 **x = 1/2 ln[(1+x)/(1-x)]

⇒ tanh-1 (1/5) = 1/2 ln[(1+(1/5))/(1 - (1/5)]

= 1/2 ln[(6/5)/(4/5)]

=1/2 ln(3/2)

Hence, tanh-1 (1/5) = 1/2 ln(3/2)

**Example 4: Find the value of sech -1 (3/8).

**Solution:

We know,

sech-1 x = ln[(1 + √(1 - x2)/x]

So, sech-1 (3/8) = ln\left[\frac{1+\sqrt{(1-(\frac{3}{8}})^{2}}{\frac{3}{8}}\right]

= ln[(8 + √(64 - 9))/3]

= ln\left[\frac{8+\sqrt{55}}{3}\right]

Hence, sech-1(3/8) = ln\left[\frac{8+\sqrt{55}}{3}\right]

**Example 5: Find the derivative of [sinh -1 (5x + 1)] 2 .

**Solution:

Let y = [sinh-1 (5x + 1)]2

Now derivative of the given function is,

dy/dx = d([sinh-1 (5x + 1)]2)/dx

= 2[sinh-1 (5x + 1)] d/dx [sinh-1 (5x + 1)

We know **d(sinh -1 x)/dx = 1/√(x 2 + 1)

= 2 [sinh-1 (5x + 1)] {1/√[(5x+1)2 + 1]} d(5x+1)/dx

= 2 [sinh-1 (5x + 1)] × {1/√(25x2 + 10x + 2)} × 5

= 10 sinh-1(5x+1)/[√(25x2+10x+2)]

Hence, the derivative of [sinh-1(5x+1)] = \frac{10sinh^{-1}(5x+1)}{\sqrt{(25x^{2}+10x+2)}} .

Example 6: If cosh x = 5/3, then prove that x = loge(5/3 + √(16/9)).

Solution:

We know that , cosh x = (e^x + e^-x )/2
Let's substiute the given value,

(5/3) = (e^x + e^(-x))/2
Multiplying both sides by 2:
10/3 = e^x + e^(-x)

Now, let's use the quadratic formula to solve for e^x:
e^x = (10/3 ± √((10/3)^2 - 4(1)))/2
Simplifying:

e^x = (10/3 ± √(16/9))/2

e^x = 5/3 ± √(16/9)

Now, taking Logarithm from both side,
x = loge(5/3 + √(16/9))

**Example 7: Prove that sinh-1(tanx) = cosh-1(secx)

**Solution:

Let's start with the definition of sinh-1.

sinh-1 (tan x) = loge(tan x + √(tan^2 x + 1))

Now, let's use the trigonometric identity:

tan^2 x +1 = sec^2 x

Substituting this into previous equation,
sinh-1 (tan x) = loge(tan x + √(sec^2 x )

Now, let's simplify

sinh-1 (tan x) = loge(√(sec^2 x-1 ) + sec x )

This is equal to cosh-1 (sec x).

**Example 8: Find the value of coth-1 (7/3).

**Solution:

We know that, coth-1 x = (1/2) * loge((x + 1)/(x - 1)).
Let's substitute the given value:

coth-1(7/3) = (1/2)*loge((7/3+1)/(7/3-1)
Simplifying:

coth-1 (7/3) = (1/2) * loge(10/4)

coth-1 (7/3) = (1/2) * loge(5/2)

**Example 9: Find the value of csch-1 (2/5).

**Solution:

We know that csch-1 x = loge(1/x + √(1/x^2 + 1)).
Let's substitute the given value:

csch-1 (2/5) = loge(5/2 + √(25/4 + 1))

Simplifying:
csch-1 (2/5) = loge(5/2 + √(29/4))

**Example 10: Find the derivative of [cosh-1 (2x + 1)]^2.

**Solution:

Let's start with the chain rule:

d/dx [cosh-1 (2x + 1)]^2 = 2 * [cosh-1 (2x + 1)] * d/dx [cosh-1 (2x + 1)]

Now, let's find the derivative of cosh-1:

d/dx [cosh-1 (2x + 1)] = 1/√((2x + 1)^2 - 1) * d/dx (2x + 1)

Simplifying:

d/dx [cosh-1 (2x + 1)] = 2/√(4x^2 + 4x)

Now, let's substitute this back into the original equation:

d/dx [cosh-1 (2x + 1)]^2 = 2 * [cosh-1 (2x + 1)] * 2/√(4x^2 + 4x)

Simplifying:

d/dx [cosh-1 (2x + 1)]^2 = 4 * [cosh-1 (2x + 1)] / √(4x^2 + 4x)

Summary

Hyperbolic functions are commonly applied today, thus inverse hyperbolic functions are also necessary to define the angle for hyperbolic functions given its value. They are the ‘‘hypotenuse is half’’, ‘ED’, ‘odd-number constant’ functions and are the hyperbolic sine, cosine, tangent, and their reciprocals which have explicit formulas and domains. All these functions are actively used in engineering, physics, and other branches of advanced mathematics. They have to be understood as well as be manageable so that one can work on different equations dealing with hyperbolic functions. Thus, the given information testifies to the possibility of learning inverse hyperbolic functions and developing the skills needed to solve real-life issues.

Practice Problems

  1. Evaluate Sinh⁻¹(1).
  2. Find the exact value of cosh⁻¹(5).
  3. Solve the equation: tanh(x) = 0.7.
  4. Simplify: sinh⁻¹(2sinh(x)).
  5. Prove that coth⁻¹(x) = ½ ln[(x + 1)/(x - 1)].
  6. Find the value of sech⁻¹(0.5).
  7. Solve for x: sinh⁻¹(x) = 2.
  8. Evaluate csch⁻¹(2).
  9. Simplify the expression: tanh⁻¹(tanh(3x)).
  10. Prove that cosh⁻¹(x) + sinh⁻¹(x) = ln(2x).

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