Find the range of $ y = \frac {\tan 3x}{\tan x} $ $$ y = \frac {\tan 3x}{\tan x} $$
Now this is what I did :
$$ y = \frac {3\tan x - \tan^3 x}{\tan x(1 - 3\tan^2 x)} $$
$$ y = \frac {3 - \tan^2x}{1 - 3\tan^2x} $$
but I was not able to do further than this, I tried converting it into $sine$ & $cosine$ form but got stuck. I think it will involve using concepts of limit and continuity but I have not been taught that yet. Any help will be appreciated :)
 A: If you multiply your numerator and denominator by $\cos^2 x$ and use $\sin^2 x=1-\cos^2 x$, the expression becomes $$\frac {3-4\cos^2 x}{1-4\cos^2 x}$$
Now set $z=4\cos^2 x$ and note that $0\le z \le 4$ and the function becomes $$\frac {z-3}{z-1}=1-\frac 2{z-1}$$
You should be able to sketch that to see what is happening - the proof will then be easy.

If $z\gt 1$ then $\frac 2{z-1}\gt 0$ and is decreasing with increasing $z$. For this without calculus suppose $w\gt z\gt 1$ then  $\frac 2{z-1}-\frac 2{w-1}=\frac {2(w-z)}{(z-1)(w-1)}\gt 0$. It can clearly be made as large as we like by choosing $z$ close to $1$, and for a decreasing function the minimum value will occur when $z$ is as large as possible and will therefore be $\frac 23$.
Since we are subtracting this term from $1$ the signs are reversed and for $z\gt 1$ the range of values is $(-\infty, \frac 13]$.
When $z\lt 1$ a similar analysis shows the range to be $[3,\infty)$

Note therefore that your original $y$ cannot take values in the open interval $(\frac 13, 3)$. This is because $0\le 4\cos^2 x\le 4$ - if the value of $z=4\cos^2 x$ were unconstrained you would be able to obtain every real value except $y=1$.
A: $\tan x$ can take any value in $(-\infty,\infty)$. So you find the range for $\frac{3-x}{1-3x}$ for $x\in[0,+\infty)$.
A: i have found that
$\frac{\tan(3x)}{\tan(x)}=\frac{1+2\cos(2x)}{-1+2\cos(2x)}$ and now you can set $t=\cos(2x)$ with $|t|\le 1$. Now you can apply calculus.
