# Value of a fraction

It it true that is ${a^2+c^2\over b^2+d^2}=1$ for $ad-bc=1$?

I tried substituting in $a={1-bc\over d}$ but it is still a mess.

(How do you ask Wolfram Alpha a question like this where we ask it to calculate something with an imposed condition?)

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Take $a = 2$ and $b = c = d = 1$. –  TMM May 13 '12 at 14:08
What is true is that $(a^2+c^2)(b^2+d^2)=(ad-bc)^2+(ab+cd)^2$. –  anon May 13 '12 at 14:15

It's not true. Try it for $a = 2$ and $b = c = d = 1$:
$$ad - bc = 2\times1 - 1\times1 = 1$$
$$\frac{a^2+c^2}{b^2+d^2} = \frac{2^2+1^2}{1^2+1^2} = \frac{5}{2} \ne 1$$
As for WA, you can use the FullSimplify function to simplify an expression given some assumptions. Here is an example.
Or even simpler: $b=c=0$, so $a={1 \over d}$ and ${{a^2+c^2}\over {b^2+d^2}}={{a^2} \over {d^2}} = {1 \over {d^4}}$ –  David Lewis May 13 '12 at 15:24