Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$ Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$
I tried:
$x=\sin\alpha$ and $y=\cos\beta$
$\sqrt{(1-x^2)(1-y^2)}=\sqrt{\cos^2\alpha\sin^2\beta}$ but if I write $\sqrt{\cos^2\alpha\sin^2\beta}=\cos\alpha \sin\beta$, it's not true because 
$\cos\alpha \sin\beta$ can be negative. 
Can someone give me an idea?

\begin{align}
& \left| xy-\sqrt{(1-x^2)(1-y^2)}\,\right| \\[10pt]
= {} & \left| \sin\alpha\cos\beta - \left| \cos\alpha\sin\beta \right| \, \right| \\[10pt]
= {} & \begin{cases} \left| \sin\alpha\cos\beta - \cos\alpha \sin\beta\right| = \left|\sin(\alpha-\beta){} \right| & \text{if } \cos\alpha>0\ \&\ \sin\beta>0, \\[5pt]
\left| \sin\alpha\cos\beta + \cos\alpha\sin\beta \right| = \left| \sin(\alpha+\beta) \right| & \text{if } \cos \alpha<0\ \&\ \sin\beta>0, \\[5pt]
\text{and similarly in the other two cases.}
\end{cases}
\end{align}
