(sinx)/(1+cosx)=(1-cosx)/(sinx)
simplify so RS=LS
You can also use $\displaystyle \frac{a}{b}=\frac{c}{d} \Leftrightarrow a\cdot d=c\cdot b$
That means for this exercice:
$\displaystyle \frac{\sin(x)}{1+\cos(x)}=\frac{1-\cos(x)}{\sin(x)} \Leftrightarrow \sin^2(x)=[1-\cos(x)]\cdot[1+\cos(x)] \Leftrightarrow \sin^2(x)=1-\cos^2(x) \Leftrightarrow \sin^2(x)=\sin^2(x)$