Question

How do you prove $\operatorname{Arccosh}\left( x \right)=\ln \left( x+{{\left( {{x}^{2}}-1 \right)}^{\dfrac{1}{2}}} \right)$ ?

Answer 1

Hint: To solve these questions the concept of hyperbolic functions should be familiar and how to expand them to solve the questions. After taking the hyperbolic functions into consideration and applying their formula, just simplify the expression by using various algebraic identities.

Complete step by step answer:
Given to prove:
$\operatorname{Arccosh}\left( x \right)=\ln \left( x+{{\left( {{x}^{2}}-1 \right)}^{\dfrac{1}{2}}} \right)$
Now, let $p={{\cosh }^{-1}}x$ then by definition of hyperbolic functions it can be expanded and written as,
$\Rightarrow x=\cosh p=\dfrac{{{e}^{p}}+{{e}^{-p}}}{2}$
Simplifying the above expression and substituting the value in the equation, we get,
$\Rightarrow 2x={{e}^{p}}+{{e}^{-p}}$
Simplifying the above equation and transposing all the expressions on the left-hand side of the equation, we get,
$\Rightarrow {{e}^{p}}-2x+\dfrac{1}{{{e}^{p}}}=0$
Taking L.C.M. of the fractions and simplifying the expression, we get,
$\Rightarrow {{e}^{2p}}-2x{{e}^{p}}+1=0$
Now, let us take $r={{e}^{p}}$ and substitute it in the above equation to get,
$\Rightarrow {{r}^{2}}-2xr+1=0$
To solve the above quadratic equation, use the quadratic formula and find the values of$r$,
$\Rightarrow r=\dfrac{2x\pm \sqrt{4{{x}^{2}}-4}}{2}$
Replacing $r$by ${{e}^{p}}$in the above expression, we get,
$\Rightarrow {{e}^{p}}=\dfrac{2x\pm \sqrt{4\left( {{x}^{2}}-1 \right)}}{2}$
Now, solve the expression under the root and transpose the denominator of the fraction to the L.H.S of the equation to get,
$\Rightarrow 2{{e}^{p}}=2x\pm 2\sqrt{{{x}^{2}}-1}$
Solving the above expression further and simplifying it, we get,
$\Rightarrow {{e}^{p}}=x\pm \sqrt{{{x}^{2}}-1}$
Taking logarithm on both the sides of the equation,
$\Rightarrow \ln{{e}^{p}}=\ln \left( x\pm \sqrt{{{x}^{2}}-1} \right)$
$\Rightarrow p=\ln \left( x\pm \sqrt{{{x}^{2}}-1} \right)$
Replacing the value of $p$in the above equation we get,
$\Rightarrow {{\cosh }^{-1}}x=\ln \left( x\pm \sqrt{{{x}^{2}}-1} \right)$
Hence, it is proved that $\operatorname{Arccosh}\left( x \right)=\ln \left( x+{{\left( {{x}^{2}}-1 \right)}^{\dfrac{1}{2}}} \right)$ .

Note: In such questions, it becomes necessary to remember the values of the hyperbolic trigonometric functions that are mentioned in the question. While solving these questions keep in mind to carefully use signs while opening parenthesis. General knowledge about logarithmic and exponential rules will also help while solving these questions.