# Let A1 be the area of the region bounded by the curves $y= \sin x, y= \cos x$ and y-axis in the first quadrant. AAlso, let A2 br the area of the region bounded by the curves $y= \sin x, y= \cos x$ x-axis and $x =\frac{\pi}{2}$ in the first quadrant. Then, Option: 1 $2A_1=A_2\text{ and } A_1 +A_2=1 +\sqrt{2}$ Option: 2 $A_{1}:A_2=1:2\text{ and }A_{1}+A_2{}=1$ Option: 3 $A_1:A_2=1:\sqrt{2}\text{ and }A_1+A_2=1$ Option: 4 $A_1=A_2\text{ and } A_1 +A_{2}=\sqrt{2}$

$\\\mathrm{A}_{1}=\int_{0}^{\pi / 4}(\cos \mathrm{x}-\sin \mathrm{x}) \mathrm{d} \mathrm{x} \\ \mathrm{A}_{1}=(\sin \mathrm{x}+\cos \mathrm{x})_{0}^{\pi / 4}=\sqrt{2}-1 \\ \mathrm{~A}_{2}=\int_{0}^{\pi / 4} \sin \mathrm{x} \mathrm{d} \mathrm{x}+\int_{\pi / 4}^{\pi / 2} \cos \mathrm{x} \mathrm{d} \mathrm{x}$

$\\=(-\cos \mathrm{x})_{0}^{\pi / 4}+(\sin \mathrm{x})_{\pi / 4}^{\pi / 2} \\ \mathrm{~A}_{2}=\sqrt{2}(\sqrt{2}-1) \\ \mathrm{A}_{1}: \mathrm{A}_{2}=1: \sqrt{2}, \mathrm{~A}_{1}+\mathrm{A}_{2}=1$

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