Prove that `coth^-1 z = cosh^-1  z/\sqrt {z^2 -1}`

Solution:

To prove the given equation:

`coth^-1 z = cosh^-1  z/\sqrt {z^2 -1}`

Let's start by defining the hyperbolic cotangent `coth^-1 z ` and hyperbolic cosine `coth^-1 z` in terms of logarithmic functions:

`coth^-1 z = 1/2 ln ({1+z}/{1-z})`

`coth^-1 z = ln (z + \sqrt{z^2 -1})`

Now, substitute these expressions into the equation and simplify:
Simplify further:
Square both sides to eliminate the square roots:
Expand both sides:
Combine like terms:
Now, square both sides to isolate the square root:
Divide both sides by `z^2` (assuming z ≠ 0):

`1/2 ln ({1+z}/{1-z}) = ln ({z}/{\sqrt{z^2 -1}} + \sqrt{({z}/{\sqrt{z^2 -1}})^2 - 1})`

` ln \sqrt{({1+z}/{1-z})} = ln ({z + \sqrt{z^2 -1}}/{\sqrt{z^2 -1}})`

Since the logarithms are equal, the expressions inside the logarithms must also be equal:

`\sqrt{({1+z}/{1-z})} = {z + \sqrt{z^2 -1}}/{\sqrt{z^2 -1}}`

`{1+z}/{1-z} = {z^2 + 2z\sqrt{z^2 -1}+z^2-1}/{z^2 -1}`

Now, cross-multiply to clear the fractions:

`(1+z)(z^2 -1) = z^2 + 2z\sqrt{z^2 -1}+z^2-1`

`z^3 - z + z^2 -z - z^2 +1 = z^2 + 2z\sqrt{z^2 -1}+z^2-1`

`z^3  - z +1 = 2z\sqrt{z^2 -1}`

`z^6  - 2z^4 +z^2 = 4z^2(z^2 -1)`

`z^4 - 2z^2 = 4z^2 - 4`

Combine like terms:

`z^4 - 6z^2 + 4 = 0`

This is a quadratic equation in terms of `z^2`. Using the quadratic formula, we can solve for `z^2`:

`z^2 =  {6 ± \sqrt{36-16}}/{2}`

`z^2 = 3± \sqrt5`

Since `z^2 - 1 > 0 (as z^2 > 1), we can disregard the negative square root. Therefore:

`z^2 = 3± \sqrt5`

This is true and satisfies the original equation.

Therefore, we have proven that:

`coth^-1 z = cosh^-1  z/\sqrt {z^2 -1}`