Question

dau coroana ...daca x= 2t supra 1+t laa pătrat si t = 1-t la pătrat supra 1+t la pătrat , calculati x la patrat+ y la patrat.

Answer 1

$\displaystyle x= \frac{2t}{1+t^2} \Rightarrow x^2= \frac{4t^2}{(1+t^2)^2} . \\ \\ \\ y= \frac{1-t^2}{1+t^2} \Rightarrow y^2= \frac{(1-t^2)^2}{(1+t^2)^2}= \frac{t^4-2t^2+1}{(1+t^2)^2}. \\ \\ \\ x^2+y^2= \frac{4t^2+t^4-2t^2+1}{(1+t^2)^2}= \frac{1+2t^2+t^4}{(1+t^2)^2} = \frac{(1+t^2)^2}{(1+t^2)^2}=1.$