Slope of The Tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$ Find slope of tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$
using $$\frac{dy}{dx} = \frac{\frac{dr}{dθ}\sin \theta + r \cos \theta}{\frac{dr}{dθ}\cos \theta-r \sin \theta}$$
I got
$$\frac {14\cos \theta \sin \theta}{7 \cos^2 \theta-7\sin \theta cos \theta}$$

The curve has polar equation
$r=7 \sin t$ 
multiply both sides by $r$
$r^2=7r\sin t$
in rectangular coordinates becomes
$x^2+y^2=7y$
Equation of a circle having center $C(0,\;7/2)$ and radius $R=7/2$
The point of the circle when $t=\pi/6$ can be found intersecting the line, in polar form, $a:t=\pi/6$ with the given circle. The line $a$ in rectangular coordinates has equation 
$$y=x\tan(\pi/6)\to x=y\sqrt 3$$
plugging this in the equation of the circle we get
$3y^2+y^2=7y\to y=0;\;y=7/4$
Therefore coordinates of $P$ are $(7/4\sqrt 3,\;7/4$)
To find the slope of the tangent in $P$ we write the equation of the polar line to the circle in $P$. The polar line of a conic in a point lying on the conic is the tangent to the conic in that point.
To find the polar line substitute in the equation of the conic$^{(*)}$
$x^2\to x_Px;\;y^2\to y_Py;\;y\to \dfrac{y+y_P}{2}$
and get
$\dfrac{7}{4}x\sqrt 3+\dfrac{7}{4}y =7\,\dfrac{y+\frac{7}{4}}{2}$
$x\sqrt 3+y=2\left(y+\frac{7}{4}\right)$
$y=x\sqrt{3} -\dfrac{7}{2}$
and finally the slope $\color{red}{m=\sqrt 3}$
Ehm... maybe not so much simpler :)
$(*)$ to complete the formulas
$x\to \dfrac{x+x_p}{2};\;xy\to \dfrac{x_Py+y_Px}{2}$
$$...$$
Hope this can be useful
