Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined I appreciate the help.
My attempt: 
$$
\begin{align}
\tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\ 
&= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\
&= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin(x)}\\
&= \frac{1}{\cos(x) \sin(x)}\\
&= \frac{1}{\frac{1}{\sec(x)}\frac{1}{\csc(x)}}\\
&=\frac{1}{\frac{1}{\sec \csc}}\\
&=\frac{1}{1}\cdot \frac{\sec(x) \csc(x)}{1}\\
&= \sec(x) \csc(x)
\end{align}
$$

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.)
