For $$k \in \mathbb{R}$$, let the solutions of the equation $$\cos\left(\sin^{-1}\left(x \cot\left(\tan^{-1}\left(\cos(\sin^{-1} x)\right)\right)\right)\right) = k$$, $$0 < |x| < \frac{1}{\sqrt{2}}$$ be $$\alpha$$ and $$\beta$$, where the inverse trigonometric functions take only principal values. If the solutions of the equation $$x^2 - bx - 5 = 0$$ are $$\frac{1}{\alpha^2} + \frac{1}{\beta^2}$$ and $$\frac{\alpha}{\beta}$$, then $$\frac{b}{k^2}$$ is equal to

Using

$$\cos(\sin^{-1}x)=\sqrt{1-x^2},$$

we get

$$\tan^{-1}\left(\cos(\sin^{-1}x)\right)=\tan^{-1}\left(\sqrt{1-x^2}\right).$$

Hence,

$$\cot\left(\tan^{-1}\left(\sqrt{1-x^2}\right)\right)=\frac{1}{\sqrt{1-x^2}}.$$

Therefore,

$$\cos\left(\sin^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right)=k.$$

Using

$$\cos(\sin^{-1}t)=\sqrt{1-t^2},$$

we obtain

$$\sqrt{1-\frac{x^2}{1-x^2}}=k.$$

$$\sqrt{\frac{1-2x^2}{1-x^2}}=k.$$

Squaring both sides,

$$k^2=\frac{1-2x^2}{1-x^2}.$$

Hence,

$$x^2=\frac{1-k^2}{2-k^2}.$$

Since

$$0<|x|<\frac{1}{\sqrt2},$$

the two solutions are

$$\alpha=\sqrt{\frac{1-k^2}{2-k^2}},\qquad\beta=-\sqrt{\frac{1-k^2}{2-k^2}}.$$

Therefore,

$$\frac{\alpha}{\beta}=-1.$$

Also,

$$\frac{1}{\alpha^2}+\frac{1}{\beta^2} =\frac{2}{\alpha^2} =2\cdot\frac{2-k^2}{1-k^2}.$$

The roots of

$$x^2-bx-5=0$$

are

$$2\cdot\frac{2-k^2}{1-k^2}\quad\text{and}\quad-1.$$

Using the product of roots,

$$-2\cdot\frac{2-k^2}{1-k^2}=-5.$$

Hence,

$$2(2-k^2)=5(1-k^2).$$

$$4-2k^2=5-5k^2.$$

$$3k^2=1.$$

$$k^2=\frac13.$$

Now, using the sum of the roots,

$$b=2\cdot\frac{2-k^2}{1-k^2}-1.$$

Substituting

$$k^2=\frac13,$$

$$b=2\cdot\frac{\frac53}{\frac23}-1.$$

$$=5-1.$$

$$=4.$$

Therefore,

$$\frac{b}{k^2}=\frac{4}{\frac13}=12.$$

Hence,

$$\boxed{12}$$