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Left Side:
$$\begin{align*}
\tan\theta+\cot\theta={\sin\theta\over\cos\theta}+{\cos\theta\over\sin\theta}={\sin^2\theta+\cos^2\theta\over\cos\theta\sin\theta}
= \dfrac{1}{1\sin\theta\cos\theta}
\end{align*}$$
Right Side:
$$\begin{align*}
\dfrac{2}{\sin2\theta}=\dfrac{2}{2\sin\theta\cos\theta}=\dfrac{1}{1\cos\theta\sin\theta}
\end{align*}$$
I got it now. Thanks!
$$\tan(\theta) + \cot(\theta) = {\sin(\theta)\over \cos(\theta)} + {\cos(\theta)\over \sin(\theta)} = {\sin^2(\theta) + \cos^2(\theta) \over\cos(\theta)\sin(\theta)} = {1\over\sin(\theta)\cos(\theta)}.$$ Now avail yourself of the fact that $$\sin(2\theta) = 2\cos( \theta)\sin(\theta).$$
I'll just put together what you wrote... $$\begin{align*} \tan\theta+\cot\theta=\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\cos\theta \cdot\sin\theta} = \dfrac{1}{\cos \theta \cdot \sin \theta} \end{align*}$$
Where the penultimate inequality is what you should have written.
Can you take it from here?
