Answer –
Let the carrier wave frequency be represented by \[{{\omega }_{c}}\] and let \[{{\omega }_{s}}\]be the signal wave frequency.
Then the received signal will be given by
\[V={{V}_{1}}cos({{\omega }_{c}}+{{\omega }_{s}})t\]
Also, the instantaneous voltage of the carrier wave is given by –
\[{{V}_{in}}={{V}_{c}}cos{{\omega }_{c}}t\]
Now upon solving, we get –
$$V.{{V}_{in}}={{V}_{1}}cos({{\omega }_{c}}+{{\omega }_{s}})t.({{V}_{c}}cos{{\omega }_{c}}t)$$
$$ ={{V}_{1}}{{V}_{c}}[cos({{\omega }_{c}}+{{\omega }_{s}})t.cos{{\omega }_{c}}t]$$
$$ =\frac{{{V}_{1}}{{V}_{c}}}{2}\left[ cos({{\omega }_{c}}+{{\omega }_{s}})t+{{\omega }_{c}}t+cos({{\omega }_{c}}+{{\omega }_{s}})t-{{\omega }_{c}}t \right] $$
$$ =\frac{{{V}_{1}}{{V}_{c}}}{2}\left[ cos(2{{\omega }_{c}}+{{\omega }_{s}})t+cos{{\omega }_{s}}t \right] $$
Only the high frequency signals are allowed to pass through a low pass filter. The low frequency signal \[{{\omega }_{s}}\] is obstructed by it.
Thus, we record the modulating signal \[2{{V}_{1}}{{V}_{c}}cos{{\omega }_{s}}t\] at the receiving station which is the signal frequency.