Show that the tangents at the extremities of chord of the cardioid through pole are perpendicular. 
Using Differential Calculus, show that the tangents drawn at the extremities of any chord of the cardioid $r=a(1+\cos{\theta})$ which passes through the pole are perpendicular to each other.

Please help me to understand the exact figure of the problem. 

If the above fig is correct, please help me to solve.
 A: For all polar curves, $$x=r\cos\theta,\\y=r\sin\theta$$
We can therefore get, by differentiation, a standard expression for the gradient:$$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}$$
For the curve $r=a(1+\cos\theta)$, we have $$\frac{dr}{d\theta}=-a\sin\theta$$
With a couple of lines of simplification, we get $$\frac{dy}{dx}=-\cot\left(\frac{3\theta}{2}\right)$$
Now the line joining the pole to a point $(r\cos\theta,r\sin\theta)$ has gradient $\tan\theta$.
Therefore, two points given by $\theta_1$ and $\theta_2$ which are collinear with the pole are connected by the relationship $$\tan\theta_1=\tan\theta_2\implies\theta_1=\theta_2+n\pi$$
In this case we can set $\theta_2=\theta_1+\pi$
Therefore the gradient at $\theta_2$ is$$-\cot\left(\frac 32(\theta_1+\pi)\right)$$
This simplifies to $$\tan\left(\frac{3\theta}{2}\right)$$
Hence, by product of gradients being $-1$, the tangents are perpendicular.
A: For  curves $\mathcal C$ with polar equation $\rho=f(\theta)$, there's a standard formula for the angle $V$ of the tangent line at the point $M$ with polar coordinates $\theta, f(\theta)$ with the polar radius:
$$\tan V=\frac{\rho(\theta)}{\rho'(\theta)}.$$
So, if we denote $N$ the point with coordinates $(\theta+\pi, f(\theta+\pi))$ and $V_1$ the angle of the tangent line at $N$ with its polar radius, these tangent lines will be perpendicular if and only if $V_1=V\pm\frac\pi2$, i.e. if and only if 
$$\tan V\tan V_1=-1\iff\rho(\theta)\rho(\theta+\pi)=-\rho'(\theta)\rho'(\theta+\pi).$$
For  the present curve, this translates as
$$a^2(1+\cos\theta)(1-\cos\theta)=a^2\sin^2\theta$$
which results from well known formulae  of mid-school…
