- #1

- 8

- 0

I'm trying to prove that

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

$$\dfrac{l}{c+u}+\dfrac{l}{c-u},$$

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?!?

Last edited: