<s>[INST] Prove trig identity: $\csc \tan - \cos = \frac{\sin^2}{\cos}$ I keep hitting seeming dead-ends. 
\begin{align*}
\csc\ x \tan\ x - \cos\ x &= \left(\frac{1}{\sin\ x}\right)\left(\frac{\sin\ x}{\cos\ x}\right) - \cos\ x \\
&= \frac{\sin\ x}{(\sin\ x)(\cos\ x)} - \cos\ x \\
&= \frac{\sin\ x}{(\sin\ x)(\cos\ x)} - \frac{(\cos\ x)(\sin\ x)(\cos\ x)}{(\sin\ x)(\cos\ x)} \\
&= \frac{(\cos^2 x)(\sin\ x)}{(\sin\ x)(\cos\ x)} \\
&= \cos\ x
\end{align*}
Thank you!
 [/INST] Your method was also not wrong,
L.H.S=$\frac{sinx}{sinxcosx}$-$\frac{cosxsinxcosx}{sinxcosx}$
= $\frac{sinx-cos^2xsinx}{sinxcosx}$
= $\frac{sinx(1-cos^2)}{sinxcosx}$
= $\frac{1-cos^2}{cosx}$
= $\frac{sin^2}{cosx}$=R.H.S
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