The line $\mathrm{x}=\frac{\mathrm{\pi }}{4}$ divides the area of the region bounded by y = sinx, y = cos x and x-axis $\left(0\le \mathrm{x}\le \frac{\mathrm{\pi }}{2}\right)$ into two regions of areas ${\mathrm{A}}_1$ and ${\mathrm{A}}_2.$ Then ${\mathrm{A}}_1:{\mathrm{A}}_2=$

$y=\sin x,y=\cos xx=x=0,x=\pi /2$

$\begin{array}{r}{A}_1={\int }_0^{\frac{\pi }{4}} \cos x-\sin xdx\\ =(\sin x+\cos x{)}_0^{\frac{\pi }{2}}\\ =\sqrt{2}-1\end{array}$

$\begin{array}{r}{A}_2={\int }_{\frac{\pi }{4}}^{\pi /2} \sin x-\cos xdx\\ =(-\cos x-\sin x{)}_{\pi /4}^{\pi /2}\\ =(-1)-\left(\frac{-1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)\\ =-1\sqrt{2}=\sqrt{2}-1\end{array}$

${\mathrm{A}}_1:{\mathrm{A}}_2=\sqrt{2 } -1:\sqrt{2 } -1=1:1$