e^{sin y} cos y$${{dy} \over {dx}}$$ + e^{sin y} cos x = cos x

Put e^{sin y} = t

e^{sin y} $$\times$$ cos y$${{dy} \over {dx}}$$ = $${{dt} \over {dx}}$$

$$ \Rightarrow $$ $${{dt} \over {dx}}$$ + t cos x = cos x

I. F. = $${e^{\int {\cos x\,dx} }} = {e^{\sin x}}$$

Solution of differential equation :

$$t.{e^{\sin x}} = \int {{e^{\sin x}}.\cos x\,dx} $$

$${e^{\sin y}}.{e^{\sin x}} = {e^{\sin x}} + c$$

at x = 0, y = 0

1 = 1 + c $$ \Rightarrow $$ c = 0

$$ \therefore $$ e^{sin x + sin y} = e^{sin x}

$$ \Rightarrow $$ sin x + sin y = sin x

$$ \Rightarrow $$ sin y = 0 $$ \Rightarrow $$ y = 0

$$ \Rightarrow y\left( {{\pi \over 6}} \right) = 0,y\left( {{\pi \over 3}} \right) = 0,y\left( {{\pi \over 4}} \right) = 0$$

$$ \therefore $$ $$1 + y\left( {{\pi \over 6}} \right) + {{\sqrt 3 } \over 2}y\left( {{\pi \over 3}} \right) + {1 \over {\sqrt 2 }}y\left( {{\pi \over 4}} \right)$$

= 1 + 0 + 0 + 0 = 1