csc(2a)-cot(2a)=tan(a) csc(2a)-cot(2a)=tan(a)=1/Sn(2a)-cos(2a)/Sn(2a)=

csc(2a)-cot(2a)=tan(a)

$$\begin{array}{rll} csc(2a)-cot(2a)&=&tan(a)\\\\ \frac{1}{sin(2a)}-\frac{cos(2a)}{sin(2a)}&=&tan(a)\\\\ 1&=&tan(a)\\\\ a&=&n\pi+\frac{\pi}{4}\qquad \mbox{Where }n \in Z \end{array}$$

Is Sn meant to be sin?

Why is there a whole string of equal signs on the second one?

Yes sin I apologize. I have no idea I was asked for help on it. It doesn't make sense?

csc(2a)-cot(2a)=tan(a)=1/Sn(2a)-cos(2a)/Sn(2a)=

It looks like this is just tentative working for the submitted question.  

anyway,  If you dont understand the answer in my last past please say so and I will expand my explanation. 

Yes expand please. I am a little lost. Thank you so much!

Sorry I just posted and lost it all.  Not sure why. 

I used the the following identity

$$\\$cos^2\theta+sin^2\theta=1$\\\\ If you divide everything by $sin^2 \theta$ you end up with\\\\ $cot^2\theta+1=csc^2\theta$\\\\ $1=csc^2\theta-cot^2\theta$\\\\ this is an identity, it is true for all angles. In the case of this question the angle is 2a but it is still true.$$

Then when you get through the next bit the answer can be in the 1st or 3rd quadrant because that is where tan is positive.

think about it and if you still need more help ask again.