C3 trig. question

sin(X-A)/cos(X+A)=1

how?

if u rearrange the equation u get : sin(X-A) - cos(X+A) = 0

I just thought maybe I should write:
sin(X-A)/cos(X+A)=1
tan(X-A)=1

then try from there for a simpler approach?

sin(X-A)/cos(X+A)=1
sinXcosA-cosXsinA/cosXcosA-sinXsinA=1
(sinXcosA-cosXsinA/cosAcosX)/(cosXcosA-sinXsinA/cosAcosX)=1
(sinX/cosX-sinA/cosA)/1-tanXtanA=1
tanX-tanA/1-tanXtanA=1
tanX-tanA=1-tanXtanA

This is what I've got so far but i have no idea where to go from here

If you keep the sin(x-A) and cos(x+A) on opposite sides of the equation it makes it simpler

$\sin x \cos A - \cos x \sin A = \cos x \cos A - \sin x \sin A$

From here, rearrange so sin x and cos x are on opposite sides, and simplify from there

Yeah I know that, the problem is actually simplifying it down now, i'm unsure how to approach it, no matter how the equation is structured. only difference is that i simplified down as a 'fraction over a fraction equals 1' as opposed to what you said. it's the same thing really

It is the same thing, but my way is easier to simplify. Rearranging gives:

$\sin x \cos A + \sin x \sin A = \cos x \cos A + \cos x \sin A$

this can be further simplified to:

$\sin x (\cos A + \sin A) = \cos x (\cos A + \sin A)$

Can you see where to go from here?