Surd Problem 1

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

$a+b = \sqrt{b^{2}+1}-\sqrt{a^{2}+1} \_\_\_\_\_\_\_\_(1)$

Similarly,

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

$a+b = \sqrt{a^{2}+1}-\sqrt{b^{2}+1} \_\_\_\_\_\_\_\_(2)$

So $(1) = (2)$ ,

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

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

$a^{2} = b^{2}$

$a = b$ or $a = -b$

If $a = b$ , then from $(1)$ we have

$2a = \sqrt{a^{2}+1}-\sqrt{a^{2}+1} = 0$

$a = b = 0$ , which is contradicted to the given condition, $a$ and $b$ are non-zero.

Therefore, $a = -b$ ; and $\frac{a}{b} = -1$ .

let $a+\sqrt{a^2+1}=x, b+\sqrt{b^2+1}=\dfrac{1}{x}$ now solve for a in terms of x $\sqrt{a^2+1}=x-a\longrightarrow a^2+1=a^2 -2ax+x^2$ $2ax=x^2-1\longrightarrow a=\dfrac{x^2-1}{2x}$ solve for b in terms of x $\sqrt{b^2+1}=\dfrac{1}{x}-b\longrightarrow b^2 +1=b^2 -\dfrac{2b}{x}+\dfrac{1}{x^2}$ $\dfrac{2b}{x}=\dfrac{1}{x^2}-1=\dfrac{1-x^2}{x^2}$ $b=\dfrac{1-x^2}{2x}$ now notice $\dfrac{1-x^2}{2x}=-\dfrac{x^2-1}{2x}\longrightarrow b=-a$ now, $\dfrac{a}{b}=\dfrac{a}{-a}=\boxed{-1}$

u can also make sub. a=tanx ,b=tany .If u put t in eqn u end up with

sin((x+y)/2)=0 implies x=2npi-y.Therefore a/b=tanx/tany=tanx/-tanx = -1