Two interfering waves have the same wavelength, frequency, and amplitude. They are traveling in the same direction but are $90^o$ out of phase. Compared to the individual waves, the resultant wave will have the same.
Diffcult
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let the equation of wave be $A \sin (k x-\omega t)$ and $A \sin \left(k x-\omega t+\frac{\pi}{2}\right)$ which can also be written as $A \cos (k x-\omega t)$
so the resultant is $S=A \sin (k x-\omega t)+A \cos (k x-\omega t)$
by observing both the waves on the graph we can see that the shape of the wave remain the same but with higher amplitude
$S=\sqrt{A^{2} \sin ^{2}(k x-\omega t)+A^{2} \cos ^{2}(k x-\omega t)+2 A^{2} \sin (k x-\omega t) \cos (k x-\omega t)}$
$S=\sqrt{A^{2}+2 A^{2} \sin (k x-\omega t) \cos (k x-\omega t)}$
$\sin 2 A=2 \sin A \cos A$ and $\sin 2 A=1-2 \sin ^{2} A$
$S=A \sqrt{1+\sin (2 k x-2 \omega t)}=A \sqrt{1+1-2 \sin ^{2}(k x-\omega t)}=\sqrt{2} A \cos (k x-\omega t)$
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