r=asintheta Find the points of horizontal and vertical tangency (if any) to the polar curve.

To find a tangent line to a polar curve,r=f(theta) we regard theta as parameter and write it's parametric equations as,
x=rcos(theta)=f(theta)cos(theta)
y=rsin(theta)=f(theta)sin(theta)
We are given the polar curve r=asin(theta)
Now let's convert polar equation into parametric equation,
x=asin(theta)cos(theta)
y=asin(theta)sin(theta)=asin^2(theta)
Slope of the line tangent to the parametric curve is given by the derivative dy/dx
dy/dx=((dy)/(d theta))/((dx)/(d theta))
Let's take the derivative of x and y with respect to theta
dx/(d theta)=a[sin(theta)d/(d theta)cos(theta)+cos(theta)d/(d theta)sin(theta)]
dx/(d theta)=a[sin(theta)(-sin(theta))+cos(theta)cos(theta)]
dx/(d theta)=a[-sin^2(theta)+cos^2(theta)]
dx/(d theta)=a(cos^2(theta)-sin^2(theta))
use the trigonometric identity:cos(2theta)=cos^2(theta)-sin^2(theta)
dx/(d theta)=acos(2theta)
dy/(d theta)=a(2sin(theta)d/(d theta)sin(theta))
dy/(d theta)=a(2sin(theta)cos(theta))
Use the trigonometric identity:sin(2theta)=2sin(theta)cos(theta)
dy/(d theta)=asin(2theta)
We locate horizontal tangents by finding the points where dy/(d theta)=0 ( provided that dx/(d theta)!=0 )
and vertical tangents at the points where dx/(d theta)=0 ( provided that dy/(d theta)!=0 )
Setting the derivative of x equal to zero for locating vertical tangents,
dx/(d theta)=0
acos(2theta)=0
=>cos(2theta)=0
2theta=pi/2,(3pi)/2,(5pi)/2,(7pi)/2
=>theta=pi/4,(3pi)/4,(5pi)/4,(7pi)/4
Let's find the corresponding radius r for the above angles,
For theta=pi/4
r=asin(pi/4)=a/sqrt(2) 
For theta=(3pi)/4
r=asin((3pi)/4)=a/sqrt(2)
For theta=(5pi)/4
r=asin((5pi)/4)=-a/sqrt(2)
For theta=(7pi)/4
r=asin((7pi)/4)=-a/sqrt(2)
Now let's set the derivative of y equal to zero for locating horizontal tangents,
dy/(d theta)=0
asin(2theta)=0
=>sin(2theta)=0
=>2theta=0,pi,2pi,3pi
=>theta=0,pi/2,pi,(3pi)/2
Now, find the corresponding radius r for above angles,
For theta=0
r=asin(0)=0
For theta=pi/2
r=asin(pi/2)=a
For theta=pi
r=asin(pi)=0
For theta=3pi/2
r=asin((3pi)/2)=-a
Note: If we plot the polar curve , its a circle and it should have two horizontal and two vertical tangents. However we got four points because it depends on a, whether it's positive or negative.
For positive value of a ,
the polar curve has horizontal tangents at (0,0),(a,pi/2)
and vertical tangents at (a/sqrt(2),pi/4),(a/sqrt(2),(3pi)/4)
For negative value of a,
the polar curve has horizontal tangents at (0,pi),(-a,(3pi)/2)
and vertical tangents at (-a/sqrt(2),(5pi)/4),(-a/sqrt(2),(7pi)/4)