How Can I Prove This Complex Fraction Equation?

[spiruel]

I'm trying to prove that

$$\dfrac{l}{c+u}+\dfrac{l}{c-u}=\dfrac{2l}{\sqrt{c^2-u^2}}$$My workings so far:
$$\dfrac{l}{c+u}+\dfrac{l}{c-u},$$
put over common denominator,
$$\dfrac{l(c-u)}{(c-u)(c+u)}+\dfrac{l(c+u)}{(c-u)(c+u)},$$
$$\dfrac{l(c-u)+l(c+u)}{(c-u)(c+u)},$$
expand out,
$$\dfrac{cl-lu+cl+lu}{(c-u)(c+u)},$$
group up and cancel out,
$$\dfrac{2cl}{c^2-u^2},$$

we need to make it equal to
$$\dfrac{2l}{\sqrt{c^2-u^2}}$$
HOW?!?

 

[vela]

I'm trying to prove that

$$\dfrac{l}{c+u}+\dfrac{l}{c-u}=\dfrac{2l}{\sqrt{c^2-u^2}}$$


My workings so far:
$$\dfrac{l}{c+u}+\dfrac{l}{c-u},$$
put over common denominator,
$$\dfrac{l(c-u)}{(c-u)(c+u)}+\dfrac{l(c+u)}{(c-u)(c+u)},$$
$$\dfrac{l(c-u)+l(c+u)}{(c-u)(c+u)},$$
expand out,
$$\dfrac{cl-lu+cl+lu}{(c-u)(c+u)},$$
group up and cancel out,
$$\dfrac{2cl}{c^2-u^2},$$

we need to make it equal to
$$\dfrac{2l}{\sqrt{c^2-u^2}}$$
HOW?!?

You can't because they're not equal. It's more apparent when you rewrite each slightly:
$$\frac{2cl}{c^2-u^2} = \frac{2cl}{c^2[1-(u/c)^2]} = \frac{2l}{c}\frac{1}{1-(u/c)^2}$$ and
$$\frac{2l}{\sqrt{c^2-u^2}} = \frac{2l}{\sqrt{c^2[1-(u/c)^2]}} = \frac{2l}{c}\frac{1}{\sqrt{1-(u/c)^2}}$$

 

[PeroK]

There's no law against plugging some numbers in! If you're not sure whether two expressions are equal, try some numbers and see! l = 1, u = 1 and c = 2 proves a lack of equality here.