Explanation

(a)
The given equation of curves are
$y = \sin x$ …(i)
and $y = \cos x$ …(ii)
From Eqs. (i) and (ii), we get
$\sin{x = \cos{x\ \ \Rightarrow \ x = \frac{\pi}{4}}}$
$\therefore$ Required area $= \int_{0}^{\pi/4}{(\cos{x - \sin{x)dx}}}$
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$= \left\lbrack \sin{x + \cos x} \right\rbrack_{0}^{\pi/4}$
$= \left( \sin{\frac{\pi}{4} + \cos{\frac{\pi}{4} - \cos 0}} \right)$
$= \left\lbrack \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1 \right\rbrack$
$= \frac{2}{\sqrt{2}} - 1 = (\sqrt{2} - 1)\ sq\ unit$