Question

The ratio in which the area enclosed by the curve $y=\cos x(0\le 0\le \frac{\pi }{2})$ in the first quadrant is divided by the curve $y=\sin x$, is:

• A. $(\sqrt{2}-1):1$
• B. $(\sqrt{2}+1):1$
• C. $\sqrt{2}:1$
• D. $\sqrt{2}+1:\sqrt{2}$

Answer 1

The correct option is C

$\sqrt{2}:1$

$\text{Area}{A}_1=\left[{\int }_0^{\pi /4}\sin xdx+{\int }_{\pi /4}^{\pi /2}\cos xdx\right]=2[-\frac{1}{\sqrt{2}}+1]=(\sqrt{2}-1)\sqrt{2}=2-\sqrt{2}\text{Area}{A}_2={\int }_0^{\pi /2}\cos xdx-A_1=[\sin x{]}_0^{\pi /2}-2+\sqrt{2}$
$=1-2+\sqrt{2}=\sqrt{2}-1$
Hence, ratio is $\frac{A_1}{A_2}=\frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}-1}=\frac{\sqrt{2}}{1}$ or $1:\sqrt{2}$