[tex]y=\dfrac{\cos x}{2\sin x}=\dfrac12\cot x\implies y'=-\dfrac12\csc^2x=-\dfrac1{2\sin^2x}[/tex]
The tangent lie is horizontal whenever [tex]y'=0[/tex]. But this is never the case, since [tex]\sin^2x>0[/tex] and so [tex]y'<0[/tex] for all [tex]x[/tex], so horizontal tangents exist.
