Explanation

(c)
The given equation is
$\sin^{4}{x + \cos^{4}{y + 2 = 4\sin{x\ \cos\ y\ }}}$
$\Rightarrow ({\sin^{2}{x - 1)}}^{2} + ({\cos^{2}{y - 1)}}^{2} + 2\sin^{2}{x + 2\cos^{2}{y - 4\sin{x\cos{y = 0}}}}$
$\Rightarrow \ ({\sin^{2}{x - 1)}}^{2} + ({\cos^{2}{y - 1)}}^{2} + 2({\sin{x - \cos{y)}}}^{2} = 0$
Which is possible only when
$\sin^{2}{x - 1 = 0,\cos^{2}{y - 1 = 0,\sin{x - \cos{y = 0}}}}$
$\Rightarrow \sin^{2}{x = 1,\cos^{2}{y = 1,\sin{x = \cos y}}}$
As $0 \leq x,y \leq \frac{\pi}{2}$
We get $\sin{x = \cos{y = 1}}$
$\therefore\sin{x + \cos{y = 1 + 1 = 2}}$