$$ \text{If for some } \alpha,\beta;\; \alpha\le\beta,\; \alpha+\beta=8$$ and  $$\sec^2(\tan^{-1}\alpha)+\cosec^2(\cot^{-1}\beta)=36,$$ $$\alpha^2+\beta^2$$  is:_______

We are given $$\alpha \le \beta$$, $$\alpha + \beta = 8$$, and $$\sec^2(\tan^{-1}\alpha) + \csc^2(\cot^{-1}\beta) = 36$$. We need to find $$\alpha^2 + \beta^2$$.

Simplify the trigonometric expressions.

Recall that $$\sec^2(\tan^{-1}\alpha) = 1 + \tan^2(\tan^{-1}\alpha) = 1 + \alpha^2$$

And $$\csc^2(\cot^{-1}\beta) = 1 + \cot^2(\cot^{-1}\beta) = 1 + \beta^2$$

(Using the identities $$\sec^2\theta = 1 + \tan^2\theta$$ and $$\csc^2\theta = 1 + \cot^2\theta$$)

Set up the equation.

$$(1 + \alpha^2) + (1 + \beta^2) = 36$$

$$\alpha^2 + \beta^2 + 2 = 36$$

$$\alpha^2 + \beta^2 = 34$$

Verification:

$$\alpha + \beta = 8$$ and $$\alpha^2 + \beta^2 = 34$$

$$(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2 = 64$$

$$2\alpha\beta = 64 - 34 = 30$$, so $$\alpha\beta = 15$$

$$\alpha$$ and $$\beta$$ are roots of $$t^2 - 8t + 15 = 0$$, giving $$t = 3$$ or $$t = 5$$.

Since $$\alpha \le \beta$$: $$\alpha = 3, \beta = 5$$. ✓

The answer is 34.