Explanation

(c) Given, $y_{1} = A\sin(kx - \omega t)$
$y_{2} = A\cos{(kx - \omega t)or\ y_{2} = a\sin\left( kx - \omega t + \frac{\pi}{2} \right)}$
Phase difference of two waves=π/2
∴resultant amplitude
$R = \sqrt{A^{2} + A^{2} + 2AAcos\ \Phi}$
$= \sqrt{A^{2} + A^{2} + 2A^{2}\cos\frac{\pi}{2}}$
$= \sqrt{A^{2}}\ \ \ \ \ \ \ \ \ \left( \because\cos{\frac{\pi}{2} = 0} \right)$
$R = \sqrt{2}A$