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.
 A: 
$1.  \ \cot^3A = \dfrac{1}{\tan^3A}\\$
$2.  \ \dfrac{1-\cot A}{\tan A } = \dfrac1{\tan A}-\dfrac{1}{\tan^2A}\\$
$3. \  \dfrac{1-\tan A}{\cot A} = \tan A - \tan^2A$

So, the RHS becomes,
$\begin{align}\dfrac{\tan^3A + \tan A - \tan^2A}{\dfrac1{\tan A}-\dfrac{1}{\tan^2A} +\dfrac{1}{\tan^3A}} &= \dfrac{\tan^3A + \tan A - \tan^2A}{\dfrac{1}{\tan^3A}\left(\tan^2A-\tan A+1\right)} \\&= \tan^3A\dfrac{\tan A(\tan^2A+1-\tan A)}{\tan^2A+1-\tan A} \\&= \tan^4A\end{align}$
A: Let $t=\tan(A)$.
Multiply both sides by rhs denominator and let $\cot(A)=1/t$. Then lhs $=t^4(1-1/t)(1/t)+t=t^3-t^2+t$, while rhs $=t^3+(1-t)t=t^3+t-t^2$.
Therefore equality.
A: $$t^4\left(\dfrac{1 -t^{-1}}{t} +t^{-3}\right)=t^3 + \dfrac{1 - t}{t^{-1}}$$
is
$$t^3-t^2+t=t^3+t-t^2.$$
A: I'll begin by saying that when you want to prove an identity like $ L=R $  you cannot multiply both sides by any value, since then it wouldn't be an identity.
Anyway, Lets say that $L$ is LHS, and $R$ is RHS. Let $t=\tan(A)$. We can then say that $\frac{1}{t}=\cot(A)$.
Therefore, RHS becomes
$$ R = \frac{t^3 + \frac{1-t}{\frac{1}{t}}}{\frac{1-\frac{1}{t}}{t} + \frac{1}{t^3}} $$
Simplifying the top and bottom (separately) shows us that
$$ R = \frac{t^3-t^2+t}{\frac{t-1}{t^2} + \frac{1}{t^3}}$$
Multiply the top and bottom by $ t^3 $, then
$$ R = \frac{t^3(t^3-t^2+t)}{t(t-1)+1} = \frac{t^4(t^2-t+1)}{t^2-t+1} = t^4 $$
Then, substituting back $ t=\tan(A) $, we get
$$ R = \tan^4(A) = L $$
Like we wanted.
