<s>[INST] 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
$$\frac {14\cos \theta \sin \theta}{7 \cos^2 \theta-7\sin \theta cos \theta}$$
 [/INST] Your derivation seems to be not correct.
Indeed:
$$x=r \cos \theta$$
$$y=r \sin \theta$$
$$r=7 \sin \theta \implies dr=7 \cos \theta d\theta$$
Thus:
$$dx=\cos \theta dr - r \sin \theta d\theta=7\cos^2 \theta-7\sin^2 \theta$$
$$dy=\sin \theta dr +r \cos \theta d\theta=7\cos \theta \sin \theta+7\cos \theta \sin \theta$$
That is
$$\frac{dy}{dx}=\frac {2\cos \theta \sin \theta}{\cos^2 \theta-\sin^2 \theta}=\frac{\frac{\sqrt 3}{2}}{\frac{3}{4}-\frac{1}{4}}=\sqrt 3$$
</s>