Original question was to show that the value of $\frac{\tan x}{\tan 3x}$ wherever defined, newer lies between 1/3 and 3.
So I solved it till here
= $ \frac{3t^2-1}{t^2-3}$. Where t = tan x
Now I need to find the range of above function.
I don't have any problems if this was linear or quadratic.
$t^2$ goes from 0 to infinity
$3t^2 – 1$ goes from -1 to infinity
But this is a fraction. How can I solve it??
Also I did try y = f(t) and swapping variables and see where y is defined ( real value) But that gives t = (1/3, 3)
y = $ \frac{3t^2-1}{t^2-3}$
Best Answer
For $|t|<3$ we obtain: $$\frac{3t^2-1}{t^2-3}=\frac{1}{3}-\frac{8t^2}{3(3-t^2)}\leq\frac{1}{3}.$$ Fot $|t|>3$ see the Dr. Mathva's post.