Verified Answer :

LammettHash

\(y=\dfrac{\cos x}{2\sin x}=\dfrac12\cot x\implies y'=-\dfrac12\csc^2x=-\dfrac1{2\sin^2x}\)

The tangent lie is horizontal whenever \(y'=0\). But this is never the case, since \(\sin^2x>0\) and so \(y'<0\) for all \(x\), so horizontal tangents exist.

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