Finding the range of values that a function $f(x,t)$ can take without tedious calculations A few days ago, I found the following question in a book only with the answer:
Question : Letting $x, t$ be real numbers, then a function $f(x,t)$ is defined as 
$$f(x,t)=\frac{(2-2\cos x)t^2+4-2\cos x}{(1-2\sin x)t^2+2t+1-2\sin x}.$$
Then, find the range of values that $f(x,t)$ can take.
The answer is 
$f(x,t)\le -\frac34$ or $f(x,t)\gt0.$
This book says,"Tedious calculations are not needed.". I've been looking for a way without tedious calculations, but I'm facing difficulty. I need your help.
 A: I've just got the following:
Since 
$$f(x,t)=\frac{\frac{1-t^2}{1+t^2}-(-3+2\cos x)}{\frac{2t}{1+t^2}-(-1+2\sin x)},$$
we know that $f(x,t)$ is the slope of the line $PQ$ where
$$P\left(\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2}\right), Q\left(-1+2\sin x,-3+2\cos x\right).$$
We can see that $Q$ moves on a circle : $(x+1)^2+(y+3)^2=2^2$ and that $P$ moves on the unit circle whose center is the origin without a point $(0,-1)$.
By drawing these figures, we can get the result.
A: Here is a partial answer, a bit too long to include as comment.  We can rewrite
$$f(x,t)=\frac{(1-\cos x)(t^2+1)+1}{(\frac{1}{2}-\sin x)(t^2+1)+t}$$
Now, as the numerator is clearly positive, $0$ is not in the range.
$f(0,t)=\dfrac{2}{(t+1)^2}$, so all positive values are in the range. 
As $f(\frac{\pi}{2}, t) = -2\dfrac{t^2 + 2}{(t-1)^2}$ it is evident that all values $< -2$ are also in the range. 
So we are left to search in $[-2, 0)$ for possible inclusion in the range or to rule out. However nothings strikes immediately except perhaps "tedius calculations", so will update in case something occurs, or perhaps someone else has a better idea...
