Trignometric problem (using De Movier's Theorem) 
Ok so this question, I started out writing tan as sin and cos in the right side of the equation, simplified as much as possible and ended up with a very (sort of) fascinating equation which is

Where s = sin thetha and c= sin thetha
As you can see the denominator and numerator looks very simple where on top sin x is with the power 5 and in the bottom cos x is with the power 5.
Then I went onto applying the De Movire theorem
And then ended up with- 

As you can see the required terms sin 5 thetha and cos 5 thetha is there, but now all I need is to cancel out those two other terms in both denominator and numerator. I simply cant find a way in which those cancel off, My question is-
How do I proceed from here?? And if I had done something wrong (if those dont cancel) then where did I go wrong? Please help
 A: Using DeMoivre's Theorem and the Binomial Theorem, you get: 
$\cos 5\theta + i\sin 5\theta = (\cos \theta + i\sin \theta)^5$ 
$= \cos^5\theta + 5i\cos^4\theta\sin\theta + 10i^2\cos^3 \theta\sin^2\theta + 10i^3\cos^2 \theta\sin^3\theta + 5i^4\cos \theta \sin^4 \theta + i^5\sin^5\theta$
$= \cos^5\theta + 5i\cos^4\theta\sin \theta - 10\cos^3 \theta\sin^2\theta - 10i\cos^2 \theta\sin^3\theta + 5\cos\theta\sin^4 \theta + i\sin^5\theta$
$= (\cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4 \theta)+i(5\cos^4\theta\sin \theta - 10\cos^2 \theta\sin^3\theta + \sin^5\theta)$
Equate real and imaginary parts to get expressions for $\cos 5\theta$ and $\sin 5\theta$. Can you finish from here?
A: This may be easier if you start with the De Moivre equation
$$\cos (5\theta) + i \sin(5\theta) = (\cos \theta + i \sin \theta)^5$$
For simplicity, let $c = \cos \theta$ and $s = \sin \theta$.  Then, the RHS is simply $(c + is)^5$.  Expanding it (using the binomial coefficients and that  $i^2 = -1$) gives:
$$c^5+5 i c^4 s-10 c^3 s^2-10 i c^2 s^3+5 c s^4+i s^5$$
or 
$$(c^5 - 10 c^3 s^2 - 5cs^4) + i (5c^4s-10c^2s^3+s^5)$$
Equating the real and imaginary parts of the original equation gives
$$\cos (5\theta) = c^5 - 10 c^3 s^2 - 5cs^4$$
$$\sin (5\theta) = 5c^4s-10c^2s^3+s^5$$
Now, what is $\tan (5\theta)$ in terms of the above?
