Why the $O(t^2)$ part in $L(t) = L + t(\csc \alpha_i - \cot \alpha_i +\csc \alpha_{i+1} - \cot \alpha_{i+1}) + O(t^2)$?

Question

Let $f$ the isoperipetric quotient and $P$ an equilateral $n$-gon ( $f(P) = \frac{1}{ 4n \tan \frac{\pi}{n}} $ )

Since $f$ is scale invariant, then we shall assume that the sides of $P$ all have length equal to one. Consider the edge between two vertices $\nu_i$ and $\nu_{i+1}$, where the vertices are considered modulo $n$. Denote by $\{\gamma_i\}_{i=1}^n$ the set of interior angles of $P$, and the exterior angles $\alpha_i := \pi - \gamma_i$. Consider moving the edge between $\nu_i$ and $\nu_{i+1}$ in the direction of the outward normal to this edge. Let $t$ denote the distance the edge is translated. Then the area of the deformation is $$A(t) = A + t \cdot |\overline{\nu_i \nu_{i+1}} | \color{red}{+ O(t^2)} = A + t \color{red}{+ O(t^2)}$$ and the perimeter $$L(t) = L + t(\csc \alpha_i - \cot \alpha_i) + t(\csc \alpha_{i+1} - \cot \alpha_{i+1})\color{red}{+ O(t^2)},$$ where $A$ and $L$ are the area and the perimeter of $P$ at $t=0$.

Question : Why do we have to add the part $O(t^2)$ at this equation?

Edit

I know that $$L(t) = L(0) + t(\csc \alpha_i - \cot \alpha_i) + t(\csc \alpha_{i+1} - \cot \alpha_{i+1}) + (1 - t(\csc \alpha_i + \csc \alpha_{i+1})), $$ but I have difficulty seeing how $(1 - t(\csc \alpha_i + \csc \alpha_{i+1})) = O(t^2).$

Question : Why do we have to add the part $O(t^2)$ at the equation of perimeter?

Answer 1

The area of the purple and green regions is $st$. The area of the green regions is $t^2\tan(\theta)$, where $\theta$ is the angle at the bottom corner of the green triangles.

Thus, the area including the purple region is $$ A+st-t^2\tan(\theta)=A+st+O\!\left(t^2\right) $$

Answer 2

The area that gets added is a trapezoid. It is almost a rectangle with side lengths $|v_iv_{i+1}|$ and $ t $ except that two right triangles a missing (the triangles that are visible in your sketch). These triangles have leg lengths proportional to $ t $, thus an area proportional to $ t^2$. Note that this area is not exactly $ t^2$ because one of the lengths has not exactly length $ t $