- #1
spiruel
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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?!?
$$\dfrac{l}{c+u}+\dfrac{l}{c-u}=\dfrac{2l}{\sqrt{c^2-u^2}}$$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?!?
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