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

$frac{dr}{dtheta}=-asintheta$ for the first curve and $frac{dr}{dtheta}=asintheta$ for the second curve but i dont know how to prove them perpendicular.

Since $(x,y)=(rcostheta,rsintheta)$, the 2 curves can be given parametrically as

begin{align}
vec{r}_1 &= big(a(1+costheta)costheta,a(1+costheta)sinthetabig) \
vec{r}_2 &= big(a(1-costheta)costheta,a(1-costheta)sinthetabig)
end{align}

Their tangent vectors are

begin{align}
vec{r}_1' &= big(a(-sintheta - sin2theta),a(costheta + cos2theta) big) \
vec{r}_2' &= big(a(-sintheta + sin2theta),a(costheta - cos2theta) big)
end{align}

Then
$$ vec{r}_1' cdot vec{r}_2' = a^2(sin^2theta - sin^2 2theta) + a^2(cos^2theta - cos^2 2theta) = 0 $$

i.e. $vec{r}_1'perp vec{r}_2' $

Alternatively, you can use

$$ frac{dy}{dx} = frac{frac{dx}{dtheta}}{frac{dy}{dtheta}} = frac{frac{dr}{dtheta}costheta - rsintheta}{frac{dr}{dtheta}sintheta + rcostheta} $$

to compute the tangent slopes that way

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 :

$$frac{dy}{dx} = frac{dy(theta)}{dtheta}/frac{dx(theta)}{dtheta}= frac{r'(theta)sin theta+r(theta)cos(theta)}{r'(theta)cos theta-r(theta)sin(theta)}=frac{r'(theta)tan theta+r(theta)}{r'(theta)-r(theta)tan(theta)} .tag{1}$$

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.

The angle between tangent line and a ray from pole of a polar curve is
$$tanpsi=dfrac{r}{r'}$$
then for these curves in every $theta$ on curves
$$tanpsi_1=dfrac{r_1}{r'_1}=dfrac{a(1-costheta)}{asintheta}=tandfrac{theta}{2}$$
$$tanpsi_1=dfrac{r_2}{r'_2}=dfrac{a(1+costheta)}{-asintheta}=-cotdfrac{theta}{2}$$
therefore
$$tanpsi_1tanpsi_2=-1$$