# What is the domain of the function $\tan\theta_{1}(\theta)=\frac{\sin\theta}{\cos\theta+1}$ and what is $\theta_1^{\max}$?

I have some questions about the lecture that I took today on Physics.

Consider the cosine function defined below,

$$\cos\theta=-\frac{M_2}{M_1}$$

$$M_1$$: Mass of the first object.

$$M_2$$: Mass of the second object.

(Sorry for the physical terms, I was undecided between open this question here or physics stackexchange. I will not jump into the formulas, don't worry.)

We have a general formula to use in different situations that is defined as,

$$\tan\theta_{1}(\theta)=\frac{\sin\theta}{\cos\theta+\frac{M_1}{M_2}}$$

Now, consider that $$M_1=M_2$$. The mass ratio will be $$1$$.

$$M_1=M_2\Rightarrow \cos\theta=-\frac{M_2}{M_1}=-1$$

$$\tan\theta_{1}(\theta)=\frac{\sin\theta}{\cos\theta+1}$$

Everything is OK until here. Now, as I remember my teacher said when,

$$\tan\theta_{1}\to\infty$$

$$\cos\theta+1=0$$

$$\cos\theta=-1$$

$$\theta=\pi$$

Here, how does the function $$\tan\theta_{1}(\theta)$$ go to $$\infty$$? I mean it doesn't have a exact result when it goes to positive infinity as I know. The trigonometric tangent function $$f(x)=\tan{x}$$ doesn't approach to a exact result although. What does my teacher mean by this limit? Another question is how he found $$\theta=\pi$$. Because,

$$(\arccos)[\cos\theta=-1]$$

$$\theta=\arccos{(-1)}$$

$$\theta=\frac{\pi}{2}$$

Got stuck...

Later on that, he found

$$0\le\tan\theta_{1}\lt\infty$$

$$0\le\theta_{1}\le\frac{\pi}{2}$$

and $${\theta_{1}}^{\max}=\frac{\pi}{2}$$.

As I think the function $$\tan\theta_{1}$$ has $$-\infty\lt\tan\theta_{1}\lt\infty$$ domain. Am I wrong?

The function you discuss has domain issues when the denominator is $$0$$, or when $$\cos\theta = -1$$.
I like to think that cosine typically refers to the $$x$$ component of the unit circle. Let me explain. Given a vector of any magnitude $$m$$ going any direction $$\theta$$, the horizontal component of the vector, associated with the distance from the origin along the $$x$$ axis, can be calculated by the following: $$i = m\cos\theta$$ where $$i$$ is the $$x$$ component. When it comes to the unit circle specifically, the magnitude of the vector is always $$1$$. Or, to phrase it another way, the unit circle has a radius of one, so no matter what the angle is, the magnitude will remain constant. However, as you move the radius around the circle, the angle between the radius horizontal $$x$$ axis will shift, thus changing the value of $$x$$ component of the vector.
So consider what we were given, we need to find a theta such that $$\cos\theta = -1$$. We could rephrase this to say we need to find the angle that will yield a horizontal component of $$-1$$. If the unit circle centered at the origin has a radius of $$1$$, then the only possible place were the $$x$$ component could be $$-1$$ would be when $$\theta$$ is strictly $$180$$ degrees, or $$\pi$$ radians.
To summarize, I think you will find that the inverse cosine of $$-1$$ is actually $$\pi$$.
As to your second question, I agree that the domain of the tangent function is $$(-\infty, \infty)$$. However, when tangent is restricted between $$(0,\infty)$$ by setting theta to be in $$(0,\frac{\pi}{2})$$, this is likely for a specific purpose. This part happens to be from the origin to the end of the second half of the first period of the tangent function. Perhaps this is related specifically to the physics question?