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I appreciate the help.
My attempt:
$$ \begin{align} \tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\ &= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\ &= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin(x)}\\ &= \frac{1}{\cos(x) \sin(x)}\\ &= \frac{1}{\frac{1}{\sec(x)}\frac{1}{\csc(x)}}\\ &=\frac{1}{\frac{1}{\sec \csc}}\\ &=\frac{1}{1}\cdot \frac{\sec(x) \csc(x)}{1}\\ &= \sec(x) \csc(x) \end{align} $$
That is exactly correct! Just two things: First, $\tan,\sin,\cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $\tan x$, $\cos x$ etc rather than just $\tan$ or $\cos$.
Finally, you could save time on your proof by noticing on the fourth step that $$ \frac{1}{\cos x\sin x}=\frac{1}{\cos x}\frac{1}{\sin x}=\sec x \csc x $$
Your steps are correct, but just keep in mind that robotically converting everying into $\sin$s and $\cos$s isn't the only option available to you.
Note that $$\cot\theta = \frac{\cos\theta}{\sin\theta}=\frac{\frac{1}{\sin\theta}}{\frac{1}{\cos\theta}}=\frac{\csc\theta}{\sec\theta}$$ that $$\cot\theta\tan\theta=\frac{1}{\tan\theta}\cdot\tan\theta=1$$ and that $$\sec^2\theta=\tan^2+1$$ then $$\begin{array}{lll} \tan\theta+\cot\theta&=&1\cdot(\tan\theta+\cot\theta)\\ &=&(\cot\theta\tan\theta)(\tan\theta+\cot\theta)\\ &=&(\cot\theta)(\tan\theta(\tan\theta+\cot\theta))\\ &=&\frac{\csc\theta}{\sec\theta}(\tan^2\theta+1)\\ &=&\frac{\csc\theta}{\sec\theta}\sec^2\theta\\ &=&\sec\theta\csc\theta \end{array}$$
An alternative approach writes $t=\tan x/2$ so $$\tan x+\cot x=\frac{2t}{1-t^2}+\frac{1-t^2}{2t}=\frac{1+t^2}{1-t^2}\frac{1+t^2}{2t}=\sec x\csc x.$$
All your work is okay.
Equations last but one, last but two and last but three can be deleted and you can straight away jump to last result because you already know $ \sec,\csc $ are inverse trig functions of $\cos, \sin$.
For acute $\theta$, there's this trigonograph:
$$\sec\theta \cdot \csc\theta \;=\; 2\,|\triangle OPQ| \;=\; 1\cdot( \tan\theta +\cot\theta)$$
(Showing that this works in any quadrant is straightforward.)
