How do we prove that $\frac{\sin{2\theta} + \cos{2\theta} + 1 }{\sin{2\theta-} \cos{2\theta}+1} = \cot{\theta}$?

Question

$$\frac{\sin{2\theta} + \cos{2\theta} + 1 }{\sin{2\theta-} \cos{2\theta}+1} = \cot{\theta}$$

I stopped here -

$\frac{(2)\sin\theta\cos\theta+ (1 + (2)\cos\theta)}{(2)\sin\theta\cos\theta + (1-(2)\cos\theta) }$

Answer 1

\begin{align*} \frac{\sin(2\theta) + \cos(2\theta) + 1}{\sin(2\theta) - \cos(2\theta) + 1} & = \frac{\sin(2\theta) + 2\cos^{2}(\theta)}{\sin(2\theta) + 2\sin^{2}(\theta)} \tag{1}\\\\ & = \frac{2\cos(\theta)(\sin(\theta) + \cos(\theta))}{2\sin(\theta)(\cos(\theta) + \sin(\theta))} \tag{2}\\\\ & = \cot(\theta) \tag{3} \end{align*}

$(1)$: I applied the identities $\cos(2\theta) = \cos^{2}(\theta) - \sin^{2}(\theta) = 2\cos^{2}(\theta) - 1 = 1 - 2\sin^{2}(\theta)$.

$(2)$: I applied the identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ and pulled out $2\cos(\theta)$ as well as $2\sin(\theta)$.

$(3)$: I simplified the obtained expression.

Hopefully this helps!