Using derivatives to determine the monotonic behavior of a polar curve

Given a polar curve defined by $$r(\theta):=1+a\cos(\theta)\tag{$a \in \mathbb R$}$$

For what intervals the polar curve is increasing/decreasing?

To determine the points for which the tangent line to the curve is parallel to the $x$-axis we need to compute: $$\frac{dy}{d\theta}=\frac{d}{d\theta}\left(\left(1+a\cos\left(\theta\right)\right)\sin\left(\theta\right)\right)=2a\cos^{2}\left(\theta\right)+\cos\left(\theta\right)-a$$

Now if we let $\frac{dy}{d\theta}=0$,we will find such points:

$$\cos\left(\theta\right)=\frac{-1\pm\sqrt{1+8a^{2}}}{4a}$$

This holds as long as: $$-1\le\frac{-1\pm\sqrt{1+8a^{2}}}{4a}\le1$$ $$\iff$$ $$-1\le\frac{-1+\sqrt{1+8a^{2}}}{4a}\le1\;\;\;\text{and}\;\;\;-1\le\frac{-1-\sqrt{1+8a^{2}}}{4a}\le1$$

Easily implies:

$$-1\le\frac{-1+\sqrt{1+8a^{2}}}{4a}\le1 \iff a\in \mathbb R-\left\{0\right\}$$ $$-1\le\frac{-1-\sqrt{1+8a^{2}}}{4a}\le1 \iff a\ge1 \;\;\;\;\text{or}\;\;\;\; a\le-1$$

So from here we conclude that the curves with $a\ge1$ or $a\le-1$ have four such tangent lines parallel to the $x$ axis and the curves with $-1<a <1$ ($a \ne 0$) have two such tangent lines.

So the points for which the tangent line to the curve is parallel to the $x$ axis are in the form:

To determine the points for which the tangent line to the curve is parallel to the $y$-axis we need to compute:

$$\frac{dx}{d\theta}=\left(\left(1+a\cos\left(\theta\right)\right)\cos\left(\theta\right)\right)=-2\sin\left(\theta\right)\cos\left(\theta\right)-a\sin\left(\theta\right)$$

Now if we let $\frac{dx}{d\theta}=0$,we will find such points:

$$\theta=k\pi \tag{$k \in \mathbb Z$}$$ $$\cos\left(\theta\right)=-\frac{1}{2a}$$

So from here we conclude that all the curves with $a \ne 0$ have such tangent lines parallel to the $y$ axis.

The Limaçon is a Cardioid iff the two internal tangent lines to the curve parallel to the $x$ axis overlap,e.g:

$$-\left(1+a\left(\frac{-1-\sqrt{1+8a^{2}}}{4a}\right)\right)\sqrt{1-\left(\frac{-1-\sqrt{1+8a^{2}}}{4a}\right)^{2}}=\left(1+a\left(\frac{-1-\sqrt{1+8a^{2}}}{4a}\right)\right)\sqrt{1-\left(\frac{-1-\sqrt{1+8a^{2}}}{4a}\right)^{2}}$$

Which happens for $\left|a\right|=1$.

The question is how from these points we are able to find out the intervals that the curve is increasing/decreasing? (For functions $f(x)$ in Cartesian coordinates we calculate the derivative of $f(x)$ and we determine the behavior of the function from roots of the derivative,but how we can do the same for polar curves?)

As you did (I simplified the expressions) $$\frac{dx}{d\theta}=-\sin (\theta ) (2 a \cos (\theta )+1) \qquad \text{and} \qquad \frac{dy}{d\theta}=a \cos (2 \theta )+\cos (\theta )$$ $$\frac{dy}{dx}=\frac{\frac{dy}{d\theta} } {\frac{dx}{d\theta} }=-\frac{\csc (\theta ) (a \cos (2 \theta )+\cos (\theta ))}{2 a \cos (\theta )+1}$$

Using the tangent half-angle substitution $\theta= 2 \tan ^{-1}(t)$ and simplifying, we have $$\frac{dy}{dx}=-\frac 1{2t}\frac{(a+1)-6 a t^2+(a-1) t^4 } {(2 a+1)+(1-2 a) t^2}$$ Now, find the regions of $t$ where $\frac{dy}{dx}$ is positive or negative.