Solution:

Given equations are
$y=\cos x \ldots(1)$
And, $y=\sin x \ldots$
$\text { Required area }=\text { Area (ABLA) }+\text { area (OBLO) }$
$\begin{array}{l}
=\int_{1}^{1} x d y+\int_{0}^{1}{\sqrt{2}} x d y \\
=\int_{\frac{1}{2}}^{1} \cos ^{-1} y d y+\int_{0}^{\frac{1}{2}} \sin ^{-1} x d y
\end{array}$
Integrating by parts, we get
$\begin{array}{l}
=\left[y \cos ^{-1} y-\sqrt{1-y^{2}}\right]_{\frac{1}{\sqrt{2}}}^{1}+\left[x \sin ^{-1} x+\sqrt{1-x^{2}}\right]_{0}^{\frac{1}{\sqrt{2}}} \\
=\left[\cos ^{-1}(1)-\frac{1}{\sqrt{2}} \cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\sqrt{1-\frac{1}{2}}\right]+\left[\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\sqrt{1-\frac{1}{2}}-1\right] \\
=\frac{-\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}-1 \\
=\frac{2}{\sqrt{2}}-1 \\
=\sqrt{2}-1 \text { sq.units }
\end{array}$
As a result, the correct answer is $\mathrm{B}$.