Show that two cardioids $r=a(1+\cos\theta)$ and $r=a(1-\cos\theta)$ are at right angles. Show that two cardioids $r=a(1+\cos\theta)$ and $r=a(1-\cos\theta)$ are at right angles.

$\frac{dr}{d\theta}=-a\sin\theta$ for the first curve and $\frac{dr}{d\theta}=a\sin\theta$ for the second curve but i dont know how to prove them perpendicular.
 A: Since $(x,y)=(r\cos\theta,r\sin\theta)$, the 2 curves can be given parametrically as
\begin{align}
\vec{r}_1 &= \big(a(1+\cos\theta)\cos\theta,a(1+\cos\theta)\sin\theta\big) \\
\vec{r}_2 &= \big(a(1-\cos\theta)\cos\theta,a(1-\cos\theta)\sin\theta\big)
\end{align}
Their tangent vectors are
\begin{align}
\vec{r}_1' &= \big(a(-\sin\theta - \sin2\theta),a(\cos\theta + \cos2\theta) \big) \\
\vec{r}_2' &= \big(a(-\sin\theta + \sin2\theta),a(\cos\theta - \cos2\theta) \big)
\end{align}
Then 
$$ \vec{r}_1' \cdot \vec{r}_2' = a^2(\sin^2\theta - \sin^2 2\theta) + a^2(\cos^2\theta - \cos^2 2\theta) = 0 $$
i.e. $\vec{r}_1'\perp \vec{r}_2' $

Alternatively, you can use
$$ \frac{dy}{dx} = \frac{\frac{dx}{d\theta}}{\frac{dy}{d\theta}} = \frac{\frac{dr}{d\theta}\cos\theta - r\sin\theta}{\frac{dr}{d\theta}\sin\theta + r\cos\theta} $$
to compute the tangent slopes that way
A: Solution 1 : take a look at the following figure :

Representing the two cardioids we are working on and 2 circles  described by $z=\frac12(1+e^{i \theta})$ and $z=\frac12(i+ie^{i \theta})$.
The 2 circles are orthogonal ; the cardioids being the images of these circles by conformal transformation $z \to z^2$ (red circle gives magenta cardioid, cyan circle gives blue cardioid), the orthogonality is preserved.
Solution 2 : Sometimes, one of the best methods for studying a polar curve $r=r(\theta)$ is to come back to its equivalent parametric representation under the form 
$$x=r(\theta)\cos \theta, \ \ y=r(\theta)\sin(\theta)\tag{1}$$
In particular, (1) permits to obtain easily the polar angle made by a tangent to a curve $r=r(\theta)$ with the horizontal axis :
$$\frac{dy}{dx} = \frac{dy(\theta)}{d\theta}/\frac{dx(\theta)}{d\theta}=  \frac{r'(\theta)\sin \theta+r(\theta)\cos(\theta)}{r'(\theta)\cos \theta-r(\theta)\sin(\theta)}=\frac{r'(\theta)\tan \theta+r(\theta)}{r'(\theta)-r(\theta)\tan(\theta)} .\tag{1}$$
Besides, the condition for orthogonality of two curves at a certain intersection point $I$ is that the product of the slopes of their tangents at this point is $-1$, i.e.,
$$\frac{dy_1}{dx} \frac{dy_2}{dx}=-1$$
I think you have now all the ingredients for being able to terminate. 
A: The angle between tangent line and a ray from pole of a polar curve is 
$$\tan\psi=\dfrac{r}{r'}$$
then for these curves in every $\theta$ on curves
$$\tan\psi_1=\dfrac{r_1}{r'_1}=\dfrac{a(1-\cos\theta)}{a\sin\theta}=\tan\dfrac{\theta}{2}$$
$$\tan\psi_1=\dfrac{r_2}{r'_2}=\dfrac{a(1+\cos\theta)}{-a\sin\theta}=-\cot\dfrac{\theta}{2}$$
therefore
$$\tan\psi_1\tan\psi_2=-1$$
