1. Given

$\tan A=\frac{1}{\sqrt{x(x^2+x+1)}},\qquad \tan B=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$

and

$\tan C=\left(x^{-3}+x^{-2}+x^{-1}\right)^{1/2}, \qquad 0<A,B,C<\frac{\pi}{2}.$

We need to find $A+B$.

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2. Simplify $\tan C$

$x^{-3}+x^{-2}+x^{-1}=\frac{1+x+x^2}{x^3}=\frac{x^2+x+1}{x^3}$

So,

$\tan C=\sqrt{\frac{x^2+x+1}{x^3}}=\frac{\sqrt{x^2+x+1}}{x^{3/2}}.$

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3. Use the formula for $\tan(A+B)$

$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}.$

Now,

$\tan A=\frac{1}{\sqrt{x}\sqrt{x^2+x+1}}, \qquad \tan B=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}.$

So,

$\tan A+\tan B=\frac{1}{\sqrt{x}\sqrt{x^2+x+1}}+\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$

Taking common denominator:

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

Also,

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

$=\frac{1}{x^2+x+1}.$

Hence,

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

Therefore,

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

Cancel $(x+1)$:

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

$=\frac{\sqrt{x^2+x+1}}{x\sqrt{x}}=\frac{\sqrt{x^2+x+1}}{x^{3/2}}.$

Thus,

$\tan(A+B)=\tan C.$

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4. Use the angle range

Since $0<A,B<\frac{\pi}{2}$, we have

$0<A+B<\pi.$

Also $0<C<\frac{\pi}{2}$.

Because $\tan(A+B)=\tan C$, the possibilities are

$A+B=C \quad \text{or} \quad A+B=\pi+C, \quad \text{etc.}$

But $A+B<\pi$, so $\pi+C$ is impossible. Also among the given options, if $A+B=\pi-C$, then

$\tan(A+B)=\tan(\pi-C)=-\tan C,$

which is not possible since we got $\tan(A+B)=\tan C$.

Hence,

$A+B=C.$

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5. Check options

• A: $C$ ✔️
• B: $\pi-C$ ✖️
• C: $2\pi-C$ ✖️
• D: $\frac{\pi}{2}-C$ ✖️

So the correct option is A.