Answer
$$y'\left( x \right) = \cot \left( {y - 1} \right)\cot x$$
Work Step by Step
$$\eqalign{
& \sin x\cos \left( {y - 1} \right) = \frac{1}{2} \cr
& {\text{use implicit differentiation differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\sin x\cos \left( {y - 1} \right)} \right] = \frac{d}{{dx}}\left[ {\frac{1}{2}} \right] \cr
& {\text{use product rule in the left side}} \cr
& \sin x\frac{d}{{dx}}\left[ {\cos \left( {y - 1} \right)} \right] + \cos \left( {y - 1} \right)\frac{d}{{dx}}\left[ {\sin x} \right] = \frac{d}{{dx}}\left[ {\frac{1}{2}} \right] \cr
& {\text{solving derivatives}} \cr
& \sin x\left[ { - \sin \left( {y - 1} \right)} \right]y'\left( x \right) + \cos \left( {y - 1} \right)\left( {\cos x} \right) = 0 \cr
& {\text{solving the equation for }}y'\left( x \right) \cr
& - \sin x\sin \left( {y - 1} \right)y'\left( x \right) = - \cos \left( {y - 1} \right)\left( {\cos x} \right) \cr
& y'\left( x \right) = \frac{{\cos \left( {y - 1} \right)\left( {\cos x} \right)}}{{\sin \left( {y - 1} \right)\sin x}} \cr
& {\text{using trigonometric identities}} \cr
& y'\left( x \right) = \cot \left( {y - 1} \right)\cot x \cr} $$
