# Thread: Prove an Equation involving Square Roots

1. ## Prove an Equation involving Square Roots

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

fankyou.

2. Originally Posted by BabyMilo
Proof $
\frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } = \sqrt{a^-1}
$

fankyou.
I believe that there is a horrible typo in the question: it should be $\frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } =\sqrt{a^2-1}$

I don't know if there is a shorthand, but I would consider writing the LHS as:
$\frac{1}{2}\sqrt{\frac{(a+1)(a-1)^2}{a-1}} +\frac{1}{2}\sqrt{\frac{(a-1)(a+1)^2}{a+1}}$

Take it from here.

Edit: I see you've corrected the typo in the OP

3. I was trying to correct it, but my internet is so slow.
Thank you for your solution

4. Hello, BabyMilo!

Another approach . . .

$\text{Prove: }\;\frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } \;=\; \sqrt{a^2-1}$

We have: . $\frac{a-1}{2}\cdot\frac{(a+1)^{\frac{1}{2}}}{(a-1)^{\frac{1}{2}}} + \frac{a+1}{2}\cdot\frac{(a-1)^{\frac{1}{2}}}{(a+1)^{\frac{1}{2}}}$

. . . . . . . $=\;\frac{(a-1)^{\frac{1}{2}}(a+1)^{\frac{1}{2}}}{2} + \frac{(a+1)^{\frac{1}{2}}(a-1)^{\frac{1}{2}}}{2}$

. . . . . . . $=\;\frac{[(a-1)(a+1)]^{\frac{1}{2}}}{2} + \frac{[(a+1)(a-1)]^{\frac{1}{2}}}{2}$

. . . . . . . $=\;\frac{(a^2-1)^{\frac{1}{2}}}{2} + \frac{(a^2-1)^{\frac{1}{2}}}{2}$

. . . . . . . $=\;(a^2-1)^{\frac{1}{2}}$

. . . . . . . $=\;\sqrt{a^2-1}$

5. ## Equation involving square roots.

Originally Posted by BabyMilo
Proof $
\frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } = \sqrt{a^2-1}
$

fankyou.
L.H.S:-
$\frac{a-1}{2} \sqrt{\frac{a+1}{a-1}} + \frac{a+1}{2}\sqrt{\frac{a-1}{a+1}}$

$= \sqrt{\frac{(a+1)(a-1)(a-1)}{(a-1)4} } + \sqrt{\frac{(a-1)(a+1)(a+1)}{(a+1)4}}$

$= \sqrt{\frac{a^2-1}{4}} + \sqrt{\frac{a^2-1}{4}} = 2 \times \frac{1}{ 2}\sqrt{a^2-1}$

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

L.H.S. = R.H.S

I hope you have understood. All the best.