Simplify expression help

sin(2pi -a)tan[(pi/2) +a]
--------------------------------------… -- = Result: 1
cos(2pi +a)tan[(3pi/2)-a]tan(pi+a)

I dunno how to simplify this, maybe something like this:
Top:
=sin(2pi -a)tan[(pi/2) +a]
=[sin(2pi)*cos(a) -cos(2pi)*sin(a)]*{[tan(pi/2) +tan(a)] /[1 -tan(pi/2)*tan(a)]}
=[0*cos(a) -1*sin(a)] * {[1+tan(a)] /[1 -0*tan(a)]}
=-sin(a) * [1 +tan(a)]

I'm not even sure this is correct, anybody please enlighten me. I just look at it and everything goes crazy in my vision! I'm confused!

Thank you!

look dont get confused
step by step
sin(2π-a)=-sin(a)
tan(0.5π-a)=-cot(a)
cos(2π+a)=cos(a)
tan(1.5π-a)=cot(a)
tan(π+a)=tan(a)
find these by supplementary and complementary angles, by symetry
so
-sin(a)[-cot(a)]/cos(a)cot(a)tan(a)
simplify - and cota
sin(a)/cos(a)tan(a)
sin/cos is tan
so
tan(a)/tan(a)
=
1

You need to be aware of things like tan(pi/2-a)=cota, tan(pi+a)=tan
sin(2pi-a)=-sina, tan(pi/2+a)=-cota.
You can get these from a suitable diagram showing 4 quadrants. Positive
angles are anticlockwise rotations about the origin starting along Ox
and negative rotations are clockwise. Have alpha an acute angle.
Look up definitions of sin, cos etc.
You can always check your answer using a calculator with a =pi/6.
The answer of 1 is correct.
I hope the craziness has subsided.

first get the values individually
Numerator
sin(2pi -a)= sina
tan[(pi/2) +a]= cota
Denominator
cos(2pi +a)=cosa
tan[(3pi/2)-a]=cota
tan(pi+a)=tana

now put them back to get
sina cota
--------------
cosa cota tana
After cancelling This will be =1

we can write sin(2pi -a) =-sina
tan[(pi/2) +a]=-cota
cos(2pi +a)=cosa
tan[(3pi/2)-a]=cota
tan(pi+a)=tana

then by simplifing we get
( -sina)(-cota)
-------------------------=1
(cosa)(cota)(tana)