I’ve the next method in Mathematica:

$$frac{c rho S left(textual content{Z0} cos left(frac{5 pi (2 n+1) x}{6 a+10 L}proper)+i c rho S sin left(frac{5 pi (2 n+1) x}{6 a+10 L}proper)proper)}{c rho S cos left(frac{5 pi (2 n+1) x}{6 a+10 L}proper)+i textual content{Z0} sin left(frac{5 pi (2 n+1) x}{6 a+10 L}proper)}$$

Code:

(c S [Rho] (Z0 Cos[(5 (1 + 2 n) [Pi] x)/(6 a + 10 L)] + 
   I c S [Rho] Sin[(5 (1 + 2 n) [Pi] x)/(6 a + 10 L)]))/(
c S [Rho] Cos[(5 (1 + 2 n) [Pi] x)/(6 a + 10 L)] + 
 I Z0 Sin[(5 (1 + 2 n) [Pi] x)/(6 a + 10 L)])

I wish to manipulate it in Mathematica by dividing the numerator and denominator by $cos left(frac{5 pi (2 n+1) x}{6 a+10 L}proper)$ in order that I get the identical expression by way of two Tan features. Is there a simple approach to do that? I’ve had a very good look within the docs and can not seem to discover something!