Question

a=x+root (x^2+1),then show that x=1/2(a-a^-1)​

Answer 1

Answer:

Step-by-step explanation:

Given :

$a=x+\sqrt{x^2+1}$

Then, prove that,

$x = \frac{1}{2}(a - a^{-1})$

Solution :

As Given,

$a=x+\sqrt{x^2+1}$

⇒ $\frac{1}{a}=\frac{1}{x+\sqrt{x^2+1}}$

By taking conjugate,

⇒ $\frac{1}{a}=\frac{1}{x+\sqrt{x^2+1}}\times\frac{x-\sqrt{x^2+1}}{x-\sqrt{x^2+1}}$

⇒ $\frac{1}{a} = \frac{x-\sqrt{x^2+1}}{(x+\sqrt{x^2+1})(x-\sqrt{x^2+1})}$

⇒ $\frac{1}{a} = \frac{x-\sqrt{x^2+1}}{(x)^2-(\sqrt{x^2+1})^2}$

⇒ $\frac{1}{a} = \frac{x-\sqrt{x^2+1}}{x^2-(x^2+1)}$

⇒ $\frac{1}{a} = \frac{x-\sqrt{x^2+1}}{x^2-x^2-1}$

⇒ $\frac{1}{a} = \frac{x-\sqrt{x^2+1}}{-1}$

⇒ $\frac{1}{a} = \sqrt{x^2+1}-x$

So,

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

⇒ $a-a^{-1}=x+\sqrt{x^2+1}-\sqrt{x^2+1}+x$

⇒ $a-a^{-1}=2x$

Hence,

$\frac{1}{2}(a - a^{-1}) = \frac{2x}{2} = x$

Hence proved,.