Identity for solving trig equation $\frac{x}{\cos t} - \frac{y}{\sin t} = z$ I have the following type of equation which I wish to solve for $t$:
$$\frac{x}{\cos(t)} - \frac{y}{\sin(t)} = z$$
I have used $c^2 + s^2 = 1$ to get it into the following form:
$$x\sqrt{1-\cos^2(t)} - y \cos(t) = z \cos(t)\sqrt{1-\cos^2(t)}$$
But now I am a little stuck as to how to continue. Is there another identity, e.g. double angle formulae that I should use?
 A: First thing, a warning: $\sin(t)$ is not necessarily equal to $\sqrt{1-\cos(t)^2}$, you need $t \in [0, \pi] \pmod {2\pi}$.
As for your problem, I would suggest putting all the $\sqrt{1-\cos(t)^2}$ on the same side of the equation and the other term on the other, factor then square the whole thing. But remember, this only gives you necessary conditions (it's a $\Rightarrow$, not a $\Leftrightarrow$), therefore you need to check all the answers you may find at the end to see if they are in the right range.
A: I suspect it will be hard to find a nice expression for the solution. If you do as zulon suggests, you will get an equation of degree 4 in $C=\cos t$. Alternatively, with $T=\tan(t/2)$ you get an equation of degree 4 in $T$ (using $\cos t = (1-T^2)/(1+T^2)$ and $\sin t=2T/(1+T^2)$).
A: $$xsec(t)-ycosec(t)=z$$ $sec(t)=\sqrt{1+T^2}$ and $cosec(t)=\frac {\sqrt {1+T^2}}{T}$ where $T=tan(t)$ . Therefore $$xT\sqrt{1+T^2}+y\sqrt{1+T^2}=z\\=>\sqrt{1+T^2}=\frac{z}{xT+y}\\=>1+T^2=\frac{z^2}{x^2T^2+y^2+2xyT}\\=>x^2T^4+2xyT^3+(x^2+y^2)T^2+2xyT+y^2-z^2=0$$Now I am afraid you will have to solve the equation keeping in mind that the expression $\frac z{xT+y}$ is positive. 
