The area of the triangle formed by the normal to the curve $\mathrm{x}= {\mathrm{e}}^{\mathrm{siny}} \mathrm{at} \left(1, 0\right)$ with the coordinate axes is

$\mathrm{x}= {\mathrm{e}}^{\mathrm{siny}} \Rightarrow 1 = {\mathrm{e}}^{\mathrm{siny}}.\mathrm{cosy}.\frac{\mathrm{dy}}{\mathrm{dx}} \Rightarrow 1 = \mathrm{x}.\mathrm{cosy}.\frac{\mathrm{dy}}{\mathrm{dx}}\Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(1,0\right) } = 1$

$\therefore$slope of normal = –1

 Equation of normal is  y-o=-1(x-1)

$\therefore$  Equation of normal is  x + y -1 = 0

$\therefore \mathrm{Area} \mathrm{of} \mathrm{triangle} \mathrm{formed} \mathrm{by} \mathrm{normal} \mathrm{with} \mathrm{axes} =\frac{c^2}{2\left|ab\right|} =\frac{{\left(-1\right)}^2}{2\left|1\times 1\right|} = \frac{1}{2}$