I need to find the tangent line where $y=a+x$ (so the tangent line of $45°$) of the following line: $r=\sqrt{\sin(2\cdot\theta)}$
first thing i do i find $x(\theta)$ and $y(\theta)$
$x=\sqrt{\sin(2\cdot\theta)}\cdot \cos(\theta)$ and $y=\sqrt{\sin(2\cdot\theta)}\cdot \sin(\theta)$
now i can determine the first derivative of both
$\frac{dx}{d\theta}=\cot(2\theta)\cos(\theta)\sqrt{\sin(2\theta)}-\sin(\theta)\sqrt{\sin(2\theta)}$ and $\frac{dy}{d\theta}=\cot(2\theta)\sin(\theta)\sqrt{\sin(2\theta)}+\cos(\theta)\sqrt{\sin(2\theta)}$
now I can find $\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$
$\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\cot(2\theta)\sin(\theta)\sqrt{\sin(2\theta)}+\cos(\theta)\sqrt{\sin(2\theta)}}{\cot(2\theta)\cos(\theta)\sqrt{\sin(2\theta)}-\sin(\theta)\sqrt{\sin(2\theta)}} = \frac{\cot(2\theta)\sin(\theta)+\cos(\theta)}{\cot(2\theta)\cos(\theta)-\sin(\theta)}$
and this is where i'm stuck