If $$\tan A = \frac{1}{\sqrt{x}\sqrt{x^2+x+1}}$$, $$\tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}}$$ and $$\tan C=x^{-3}+x^{-2}+x^{-1{\frac{1}{2}}}$$, $$0 < A, B, C < \frac{\pi}{2}$$, then $$A + B$$ is equal to:

Using the identity $$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$:

Let $$k = \sqrt{x^2+x+1}$$. Then $$\tan A = \frac{1}{\sqrt{x}k}$$ and $$\tan B = \frac{\sqrt{x}}{k}$$.

$$\tan(A+B) = \frac{\frac{1}{\sqrt{x}k} + \frac{\sqrt{x}}{k}}{1 - \frac{1}{k^2}} = \frac{\frac{1+x}{\sqrt{x}k}}{\frac{k^2-1}{k^2}} = \frac{(1+x)k}{\sqrt{x}(k^2-1)}$$.

Substitute $$k^2-1 = x^2+x$$:

$$\tan(A+B) = \frac{(1+x)\sqrt{x^2+x+1}}{\sqrt{x}(x^2+x)} = \frac{(1+x)\sqrt{x^2+x+1}}{x\sqrt{x}(1+x)} = \frac{\sqrt{x^2+x+1}}{x^{3/2}}$$.

This simplifies to $$\sqrt{\frac{x^2+x+1}{x^3}} = \sqrt{x^{-1} + x^{-2} + x^{-3}}$$.

Looking at $$\tan C$$, if we assume the expression in the image (which is slightly blurry but fits the pattern) represents the square root of the sum or is equivalent to this expansion, then $$\tan(A+B) = \tan C$$.

Since $$0 < A,B,C < \pi/2$$, then $$A+B = C$$.