Official solution
\[{{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x\]
\[ \Rightarrow \] \[\int {{{dy} \over {y + 1}}} = \int {{{\left( { - \cos x} \right)dx} \over {2 + \sin x}}} \]
\[ \Rightarrow \] \[\ln \left| {y + 1} \right| = - \ln \left| {2 + \sin x} \right| + \ln c\]
\[ \Rightarrow \] \[\ln \left| {\left( {y + 1} \right)\left( {2 + \sin x} \right)} \right| = \ln c\]
As y(0) = 1
\[ \therefore \] \[\ln \left| {\left( {1 + 1} \right)\left( {2 + 0} \right)} \right| = \ln c\]
\[ \Rightarrow \] \[ \therefore \] c = 4
\[ \therefore \] \[{\left( {y + 1} \right)\left( {2 + \sin x} \right)}\] = 4
\[ \Rightarrow \] y = \[{{{2 - \sin x} \over {2 + \sin x}}}\]
Given y(\[\pi \]) = a
\[ \therefore \] y(\[\pi \]) = \[{{{2 - \sin \pi } \over {2 + \sin \pi }}}\] = 1 = a
\[{{dy} \over {dx}} = {{\left( {2 + \sin x} \right)\left( { - \cos x} \right) - \left( {2 - \sin x} \right).\left( {\cos x} \right)} \over {{{\left( {2 + \sin x} \right)}^2}}}\]
\[ \Rightarrow \] \[{\left. {{{dy} \over {dx}}} \right|_{x = \pi }} = \] 1 = b
So, (a, b) = (1, 1)