c
Considering a small area swept by a planet in time dt is, \(d A=\frac{1}{2} r \cdot(r d \theta)\)

\(\therefore\) So, rate of swept of area.

\(\frac{d A}{d t}=\frac{1}{2} r\left(r \frac{d \theta}{d t}\right)\)

We know angular momentum, \(\vec{L}=m r^2 \frac{d \vec{\theta}}{d t}\)

So, \(\frac{d A}{d t}=\frac{1}{2 m}\left(m r^2 \frac{d \theta}{d t}\right)=\frac{L}{2 m}\)

Now, a planet sweeps equal amount of area in same time interval of its revohution. (from 2 nd law) So, \(\frac{d A}{d t}=\) constant; Hence \(L\) is also constant

\(\therefore\) The \(2^{nd}\) law is nothing but a shatement of conservation of angular momentum.