prove equation equals $-1$

Question

I was wondering if it was possible to prove that
$\left( \frac{( a^2 - c^2 + (\frac{d}{c}a)^2-d^2)}{ \sqrt{(a-c)^2+(\frac{d}{c}a)-d)^2} \sqrt{(a+c)^2+(\frac{d}{c}a)+d)^2}} \right) = -1$ when $a,d,c \in [-1,1]$ and $|a|<c$, and $|b|<d$?

Attempt: By Asdrul, I know that $$\left[(a-c)^{2}+\bigg(\frac{d}{c}a-d\bigg)^{2}\right]\cdot\left[(a+c)^{2}+\bigg(\frac{d}{c}a+d\bigg)^{2}\right]=\frac{1}{c^4}( c^{2}+d^{2}) ^{2}\left( a^2-c^2\right) ^{2}$$ So applying this, I have that the above equals $\frac{c^2(a^2-c^2+(\frac{da}{c})^2-d^2)}{(c^2+d^2)|a^2-c^2|}$

score 2 · Accepted Answer · edited Nov 14 '14 at 05:08

2

Hint: note that

$$\left[(a-c)^{2}+\bigg(\frac{d}{c}a-d\bigg)^{2}\right]\cdot\left[(a+c)^{2}+\bigg(\frac{d}{c}a+d\bigg)^{2}\right]=\frac{1}{c^4}( c^{2}+d^{2}) ^{2}\left( a^2-c^2\right) ^{2}$$

edited Nov 14 '14 at 05:08

daOnlyBG

2,711

answered Nov 14 '14 at 03:14

AsdrubalBeltran

4,212

Thanks for your quick reply! Okay, so I then get $\frac{c^2((a^2-c^2)+(b^2-d^2))}{(c^2+d^2)|a^2-c^2|}$ but I am still unsure of where to go from there – Kate Nov 14 '14 at 03:20
Now see that if $$|a|<c\implies |a^2-c^2|=-(a^2-c^2)$$ – AsdrubalBeltran Nov 14 '14 at 10:03
hm, yes I see that. I'm still confused on how to cancel the numerator and denominator though. Thanks for all of your help – Kate Nov 14 '14 at 13:58
figured it out, thanks a million:) – Kate Nov 14 '14 at 15:15

prove equation equals $-1$

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