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