hello guys, i got a problem in solving this
if
$\displaystyle x=\tfrac{2ab}{{b_{}}^{2}+1}
$
calculate :
$\displaystyle \tfrac{\sqrt{a+x} - \sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x}} $
thanks
Why not just calculate the 2 relevant terms first to see what you are dealing with ?
$\displaystyle a+x=a+\frac{2ab}{b^2+1}=\frac{a\left(b^2+1\right)} {b^2+1}+\frac{2ab}{b^2+1}$
$\displaystyle =\frac{ab^2+a+2ab}{b^2+1}=\frac{a\left(b^2+2b+1\ri ght)}{b^2+1}$
$\displaystyle =\frac{a(b+1)^2}{b^2+1}$
Then
$\displaystyle \sqrt{a+x}=(b+1)\sqrt{\frac{a}{b^2+1}}$
Similarly
$\displaystyle \sqrt{a-x}=(b-1)\sqrt{\frac{a}{b^2+1}$
which gives
$\displaystyle \sqrt{a+x}-\sqrt{a-x}=2\sqrt{\frac{a}{b^2+1}}$
$\displaystyle \sqrt{a+x}+\sqrt{a-x}=2b\sqrt{\frac{a}{b^2+1}}$
Then it's quite straightforward to perform the division.