General solution of the differential equation $$\cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d} y=0$$ is

If the line $$a x+b y+c=0$$ is a normal to the curve $$x y=1$$, then

Let $$\bar{a}, \bar{b}, \bar{c}$$ be three non-zero vectors, such that no two of them are collinear and $$(\bar{a} \times \bar{b}) \times \bar{c}=\frac{1}{3}|\bar{b}||\bar{c}| \bar{a}$$. If $$\theta$$ is the angle between the vectors $$\bar{b}$$ and $$\bar{c}$$, then the value of $$\sin \theta$$ is

If $$\mathrm{f}(x)=\mathrm{e}^x, \mathrm{~g}(x)=\sin ^{-1} x$$ and $$\mathrm{h}(x)=\mathrm{f}(\mathrm{g}(x))$$, then $$\frac{\mathrm{h}^{\prime}(x)}{\mathrm{h}(x)}$$ is