Question: Before algebraic manipulation:

$$\lim\_{\theta\to-\pi/4}\frac{1+\sqrt2\sin\theta}{\cos2\theta}$$

After algebraic manipulation:

$$\lim\_{\theta\to-\pi/4}\frac1{1-\sqrt2\sin\theta}$$

Difference between the functions:
'Before' is not defined at $-\dfrac{\pi}4$ while the 'After' is.

Info I have about indeterminate forms:
It is a resultant of two 'calculus rules' 'competing' to govern the answer because of lack of 'information'

1. Using this information, please explain why the manipulation and loss of information (regarding $-\dfrac{\pi}4$) made it possible to take the limit of this expression?
2. Why does the limit of the first function take the indeterminate form?

My question is not only about this specific case, but also about cases where after getting the indeterminate form, we can multiply functions by their conjugate or simplify them and all of the sudden their limits are valid.

Try to make your answer as detailed as possible.

Answer: 
A limit is evaluated from all values in the neighborhood of the target value, and not at the target value itself. So all simplifications are legit.

For instance,

$$\lim\_{x\to1}\frac{x^2-1}{x-1}=\lim\_{x\to1}(x+1)$$ because

$$x-1\ne0\implies \frac{x^2-1}{x-1}=x+1.$$

A limit tells what the function value should be when this value is indeterminate (more precisely what it should be to preserve continuity).
One has
$$ \frac{1+\sqrt{2}\sin(\theta)}{\cos(2\theta)} = \frac{(1+\sqrt{2}\sin(\theta))(1-\sqrt{2}\sin(\theta))}{(\cos(2\theta))(1-\sqrt{2}\sin(\theta))} = \frac{1-2\sin^2(\theta)}{(1-2\sin^2(\theta))(1-\sqrt{2}\sin(\theta))} =\frac{1}{1-\sqrt{2}\sin(\theta)}$$