Prove that $\tan(x)-\tan(3x)-\tan(\pi/2-3x)+\tan(\pi/2-x)=4\cos(4x)/\sin(6x).$ I have been working on it but haven't managed to prove it. This is $(b)$ part of the question. In part $(a)$, I have proved that $\tan(y)+\cot(y)=\frac{2}{\sin(2y)}$. My steps are as follows:
\begin{equation*}
LHS=\tan(x)-\tan(3x)-\tan\left(\frac{\pi}{2}-3x\right)+\tan\left(\frac{\pi}{2}-x\right)
= \tan(x)-\tan(3x)-\cot(3x)+\cot(x)
=\tan(x)+\cot(x)-(\tan(3x)+\cot(3x))
=\frac{2}{\sin(2x)}-\frac{2}{\sin(6x)}.
\end{equation*}
And I don't know how to continue to prove it to be $\frac{4\cos(4x)}{\sin(6x)}$.
Hope you can help
 A: Since just giving you the solution will mostly just be meaningless I'll also try guide you on how to think about this question.
We have:
$$LHS = \frac{2}{\sin(2x)} - \frac{2}{\sin(6x)}$$
These sorts of trigonometry problems can be a real challenge because there's so much we can do here to change the form of this expression. With these types of problems, it's very important to look carefully at the final form you want and clues within the question so that you don't lose your way. We want:
$$LHS = \frac{4\cos(4x)}{\sin(6x)}$$
Now notice that our current $LHS$ expression has a $\sin(6x)$ as one of the denominators and for our final expression we have this on our denominator. However, when we combine the two terms we have, we are also going to have a $\sin(2x)$ on our denominator which is unwanted. This information now lets us form a plan:

*

*Combine the two terms.

*Look for a way to get $\sin(2x)$ in all the terms on our numerator.

*Cancel out the $\sin(2x)$s and rearrange for $\cos(4x)$.
Since we know $LHS=RHS$ is true, provided we can cancel out a factor of $\sin(2x)$ this will basically always work.
With that being said:


 $$\frac{2}{\sin(2x)} - \frac{2}{\sin(6x)} = \frac{2(\sin(6x)- \sin(2x))}{\sin(2x)\sin(6x)}$$
$$= \frac{2(\sin(2x)\cos(4x) + \sin(4x)\cos(2x)-\sin(2x))}{\sin(2x)\sin(6x)}$$
$$= \frac{2(\sin(2x)\cos(4x) + 2\sin(2x)\cos(2x)\cos(2x)-\sin(2x))}{\sin(2x)\sin(6x)}$$
$$= \frac{2(\cos(4x) + 2\cos^2(2x)-1)}{\sin(6x)}$$
$$\frac{2}{\sin(2x)} - \frac{2}{\sin(6x)} = \frac{2(\cos(4x) + \cos(4x))}{\sin(6x)} = \frac{4\cos(4x)}{\sin(6x)}$$
 Good luck in future questions!

A: Recall that $2\sin(a)\cos(b) = \sin(a-b)+ \sin(a+b) $. Thus, choosing $a = 2x$, $b=4x$ and dividing by $\sin(2x)$ gives
$$
2\cos(4x) =\frac{1}{\sin(2x)}\left(\sin(-2x) +\sin(6x) \right) = -1 + \frac{\sin(6x)}{\sin(2x)} 
$$
Hence
$$
\frac{2\cos(4x)}{\sin(6x)} = \frac{\frac{\sin(6x)}{\sin(2x)} -1}{\sin(6x)} = \frac{1}{\sin(2x)} - \frac{1}{\sin(6x)}
$$
as desired.
