<s>[INST] prove the trigonometry identity: prove the trigonometry identity:

$$\tan^4(A) = \frac{\tan^3(A) + \frac{1 - \tan(A)}{\cot(A)}}{\frac{1 - \cot(A)}{\tan(A)} + \cot^3(A)}$$

of course i started from the complicated side the RHS and i wrote them all into tangents but then it all messy up and I'm stuck there.
 [/INST] 
$1.  \ \cot^3A = \dfrac{1}{\tan^3A}\\$
$2.  \ \dfrac{1-\cot A}{\tan A } = \dfrac1{\tan A}-\dfrac{1}{\tan^2A}\\$
$3. \  \dfrac{1-\tan A}{\cot A} = \tan A - \tan^2A$

So, the RHS becomes,
$\begin{align}\dfrac{\tan^3A + \tan A - \tan^2A}{\dfrac1{\tan A}-\dfrac{1}{\tan^2A} +\dfrac{1}{\tan^3A}} &= \dfrac{\tan^3A + \tan A - \tan^2A}{\dfrac{1}{\tan^3A}\left(\tan^2A-\tan A+1\right)} \\&= \tan^3A\dfrac{\tan A(\tan^2A+1-\tan A)}{\tan^2A+1-\tan A} \\&= \tan^4A\end{align}$
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