Trignometric identiy (LOADS o HELP ;D )

[clearlyjunk]

Homework Statement

sin(pi +A) sec(pi-A)
--------- + ---------- = 0
cos (2pi + A) csc(2pi-A)

Homework Equations

cscx=1/sinx
secx=1/cosx
sin(x+y)=sinxcosy+cosxsiny
sin(x-y)=sinxcosy-cosxsiny
cos(x+y)= cosxcosy-sinxsiny
cos(x-y)= cosxcosy+sinxsiny

The Attempt at a Solution

LS
sin pi cosA + cosA sin pi || 1/cos pi cos A + sin pi sin A
------------------------ + --------------------------
cos 2pi cosA + sin 2pi sinA || 1/sin pi cos A - cos pi sin A


the || is a seperator of the fractions

^
if it helps, i got this one to equal zero :)

 

[Mark44]

First off, use the identities that involve the sum of pi and an angle. E.g., sin(pi + A) = -sin(A) and cos(pi + A) = -cos(A). Also, cos(2pi + A) = cos(A) and so on.

Since all of the trig functions are periodic with period 2pi, f(2pi + A) = f(A), where f represents sine, cosine, tangent (yes I know that tangent's period is pi), and all the rest.

In the work you did, you have sin(pi) and cos(2pi) and all. Simplify those to 0 and 1, respectively.

 

[abstrakt!]

Is this your equation?

$\frac {\sin (\pi+A)}{\cos(2\pi+A} \ + \ \frac{\sec(\pi-A)}{\csc(2\pi+A)} \ = \ 0$