Deriving Maclaurin series

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<ol class="abc-list" start="2"><li data-list-item-id="item-1760430167252-ya56ib33d" data-list-item-path="item-1760430167252-ya56ib33d"><div class="list-item-paragraph">An arc $PQ$ of a circle subtends an angle $\left( \pi -\theta \right)$ radians at the centre $O$. The length of chord $PQ$ is $L$ units and the area of the shaded segment $PRQ$ is $A$ square units.</div><div class="list-item-paragraph">Given that&nbsp;$\theta $ is small and $A=k{{L}^{2}}$, show that&nbsp;$2k{{\theta }^{2}}-2\theta +\pi -8k\approx 0$.<span class="inline-marks">[5]</span></div></li></ol>

An arc $PQ$ of a circle subtends an angle $\left( \pi -\theta \right)$ radians at the centre $O$. The length of chord $PQ$ is $L$ units and the area of the shaded segment $PRQ$ is $A$ square units.

Given that $\theta $ is small and $A=k{{L}^{2}}$, show that $2k{{\theta }^{2}}-2\theta +\pi -8k\approx 0$.

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