Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined

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 $$\begin{array}{lll} \tan\theta+\cot\theta&=&1\cdot(\tan\theta+\cot\theta)\\ &=&(\cot\theta\tan\theta)(\tan\theta+\cot\theta)\\ &=&(\cot\theta)(\tan\theta(\tan\theta+\cot\theta))\\ &=&\frac{\csc\theta}{\sec\theta}(\tan^2\theta+1)\\ &=&\frac{\csc\theta}{\sec\theta}\sec^2\theta\\ &=&\sec\theta\csc\theta \end{array}$$