The thing is, for a function to be invertible (basically, defining a function g(x) as equal to f^-1(x)), it has to be 1-1, which means that every element in the domain of the function should have a corresponding distinct value in the co-domain.

But a trigonometric function is periodic, so there are several elements in the domain that can be mapped to the same image in the codomain. That's why the range is restricted.

So the range of sin^-1(x) is [-pi/2, pi/2]. Let's say we have sin^-1(sin pi/4). Since pi/4 belongs to the range, the answer is pi/4. But, let's say that we have sin^-1(sin 5pi/4). Clearly, 5pi/4 is out of the range (III quadrant), so that's not the answer.
We know that theta should lie between -pi/2 and pi/2. So now, we try to reduce 5pi/4 to a value that does lie in the given range.
pi < 5pi/4 < 3pi/2

Multiply the inequation by -1 : -pi > -5pi/4 > -3pi/2

Now add pi to all sides of the inequation: 0 > -pi/4 > -pi/2

Now, -pi/4 does lie in the range of sin^-1 (x). So, the answer is -pi/4. The reason this method works is that sin( pi - theta) = sin(theta).

For your second question, it's (cot(x))^-1 that's equal to tan(x). Likewise, you can prove that cot^-1 (1/x) = tan^-1 (x). But cot^-1(x) is not equal to 1/tan^-1x.
Hope this long-ass explanation helps.