Functional Trigonometry

Note first that $\cot^{2}(\theta) + 1 = \csc^{2}(\theta),$ and thus

$f(\theta) = \dfrac{(\csc^{2}(\theta) + 4)(\csc^{2}(\theta) + 9)}{\csc^{2}(\theta)} = \csc^{2}(\theta) + 13 + \dfrac{36}{\csc^{2}(\theta)}.$

Now since $\csc^{2}(\theta)$ has no upper bound (and is always $\ge 1$ ), we know that the range of $f(\theta)$ goes to $\infty.$ For the minimum, by the AM-GM inequality we have that

$f(\theta) \ge 13 + 2*\sqrt{\csc^{2}(\theta)*\dfrac{36}{\csc^{2}(\theta)}} = 13 + 2*6 = 25.$

This minimum can be achieved when $\csc^{2}(\theta) = 6,$ and so the range of $f(\theta)$ is $\boxed{[25,\infty)}.$

We could also use some calculus. Letting $u = \csc^{2}(\theta),$ we find that

$\dfrac{df}{d\theta} = \dfrac{df}{du} \dfrac{du}{d\theta} = \dfrac{(2u + 13)u - (u^{2} + 13u + 36)}{u^{2}} * (-2) * \csc^{2}(\theta)\cot(\theta) =$

$\pm \dfrac{(u^{2} - 36)*2u\sqrt{u - 1}}{u^{2}} = \pm \dfrac{2(u^{2} - 36)\sqrt{u - 1}}{u},$

where I brought in the fact that $\cot(\theta) = \pm \sqrt{\csc^{2}(\theta) - 1}.$

So the critical points, (noting that $u = \csc^{2}(\theta) \ge 1$ ,) are $u = 1$ and $u = 6.$

At $u = \csc^{2}(\theta) = 1$ the given function takes on the value $50,$ (which turns out to be a local maximum), and at $u = 6$ the function takes on a value of $\dfrac{10*25}{6} = 25.$ We can thus conclude that the range of the given function is $\boxed{[25, \infty)}.$

Replacing $cot(θ)^2$ with the variable $x$ , we can graph the function and see the minimum value.

From the graph, the minimum value is $25$ , when $x=5$ . Since $cot(θ)^2 = x = 5$ has a defined value of $θ$ , the minimum value of the expression is $25$ . $cot(θ)^2$ can go till infinity, making the upper bound infinity.

write $\superscript {cot\theta}{2}$ as x and since x must be positive solve algebraically since i think solving with calculus will be painful ! use the condition $D$ >0 and get the answer !