I've been working on this for like half an hour now and don't seem to be getting anywhere. I've tried using the double angle identity to write the LHS with $\theta/2$ and have tried expanding the RHS as a sum. The main issue I'm having is working out how to get a $\pi$ on the LHS or removing it from the right to equate the two sides.
$$\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$$
Answer
hint:
\begin{align}
& \cos\theta = \cos^2(\theta /2) - \sin^2(\theta/2)= (\cos (\theta/2) +\sin (\theta/2))(\cos (\theta/2) - \sin (\theta/2)), \\[10pt]
& 1 \pm\sin \theta = (\cos (\theta/2) \pm \sin(\theta/2))^2.
\end{align}
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