Composition of $$\tan^{-1}x$$ with $$\tan$$

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.

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

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

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

The answer is 34.