Reasons behind Assumptions to Simplify Inverse Trigonometric Functions

Question

I have recently tried learning about Inverse Trigonometric Functions and may seem to have stumbled upon a roadblock-

While trying to simplify the following equation:

$$ \arctan \frac{(1+x^2)^{1/2}-1}{x} \qquad \text{where }x \neq0 $$ I have done my Google research most of the answers assume statements like let $x = \tan A$ or let $x = \cot A$. I am OK with it because it solves my problem but I really do not know why it works (or does it really?).

I am no expert in these kinds of problems so I would really be grateful if anyone could elaborate that it works, why it works and what assuming actually means in this context.

Edit 1: I have asked question specifically about tan inverse but I would like to know about all the symmetries that exist with the Trigonometric functions.

Edit 2: It seems that this question has nothing to do with symmetries so as pointed out in the comments I am looking for cases where Assumptions such as y = f(x) can be made

Thanks

Answer 1

I'll try to answer the question "When can I say 'Let $x = f(A)$', and why can we make those assumptions?", and I'm going to assume zero knowledge of functions here. If you know more, then it would be easier to understand.

This actually is more general than trigonometric functions, this is connected to the range of a function. The range of a function $f$ is the set of all elements that we can obtain by applying $f$ to an element. For example, the function $f(x) = x^2$ has $[0, \infty)$ as range. If you draw this function, you can notice that $[0, \infty)$ is the set you obtain if you 'squeeze' vertically this function. This is called a projection of the graph into the line $x=0$. So we can obtain the range of a function from reals to reals if you draw it and then you project it into the line $x=0$ (imagine taking the graph with both of your hands and 'pushing' it on this line).

Because the range of the function $f(x) = x^2$ is exactly $[0, \infty)$, if I give you an element $y$ from $[0,\infty)$ then there is an element $z$ such that $z^2 = y$, because that's the definition of being in the image set. So it's perfectly fine to say "Let $y = f(z) = z^2$", using this fact.

Now let's go back to your original example. You have $x$ a real number. And someone says "Let $x = \tan(A)$". They can assume that because the range of the function $\tan$ is all reals, that is, $(-\infty, \infty)$, so no matter what was $x$, there is always an angle $\theta$ such that $\tan \theta = x$. You could do this with other trigonometric functions, like $\cos$, $\sin$, but they have another ranges, in this case $[-1,1]$. When applying this you should know beforehand what is the range of the function you're working with.

Answer 2

Consider this as a change of variable. We observe that the image of $\tan \phi$, for $\phi \in (-\pi/2,\pi/2)$ is the real line.

Answer 3

Let $\arctan\dfrac{\sqrt{1+x^2}-1}x=y$

$\implies-\dfrac\pi2<y<\dfrac\pi2$

$$\implies1+x^2=(1+x\tan y)^2$$

$$\implies x=\dfrac{2\tan y}{1-\tan^2y}=?$$