The area formed by triangular shaped region bounded by the curves y = sin x, y = cos x and x = 0 is

$\sqrt{2}-1$ 1 $\sqrt{2}$ $1+\sqrt{2}$

$\sqrt{2}-1$

Area $=\left|\int\limits_0^{\pi / 4}(\sin x-\cos x) d x\right|=\sqrt{2}-1$ (∵ y = sin x and y = cos x meet when sin x = cos x ⇒ x = $\frac{\pi}{4}$) Hence (1) is the correct answer.