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Step-by-Step Solution
Step 1: Express the given tangents
We have three angles A, B, and C in the interval 0 to $ \frac{\pi}{2} $ with:
$ \tan A = \frac{1}{\sqrt{x \bigl(x^2 + x + 1\bigr)}}, \quad \tan B = \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}}, \quad \tan C = \sqrt{x^{-3} + x^{-2} + x^{-1}}. $
Step 2: Compute $ \tan A \times \tan B $
$ \tan A \times \tan B
= \frac{1}{\sqrt{x(x^2 + x + 1)}}
\times \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}}
= \frac{\sqrt{x}}{\sqrt{x(x^2 + x + 1)} \, \sqrt{x^2 + x + 1}}.
$
Noting that $ \sqrt{x} \times \sqrt{x} = x $ and $ (x^2 + x + 1) \times (x^2 + x + 1) = (x^2 + x + 1)^2 $, we simplify to get:
$ \tan A \times \tan B
= \frac{\sqrt{x}}{ \sqrt{x} \,(x^2 + x + 1) }
= \frac{1}{x^2 + x + 1}.
$
Step 3: Compute $ \tan A + \tan B $
$ \tan A + \tan B
= \frac{1}{\sqrt{x} \,\sqrt{x^2 + x + 1}}
+ \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}}.
$
Take the common denominator $ \sqrt{x} \,\sqrt{x^2 + x + 1} $:
$ \tan A + \tan B
= \frac{1 + x}{\sqrt{x} \,\sqrt{x^2 + x + 1}}.
$
Step 4: Use the tangent addition formula
Recall the formula for the tangent of a sum of two angles:
$ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \,\tan B}. $
Substitute the expressions for $ \tan A + \tan B $ and $ \tan A \,\tan B $:
$
\tan(A + B)
= \frac{\frac{1 + x}{\sqrt{x} \,\sqrt{x^2 + x + 1}}}
{1 - \frac{1}{x^2 + x + 1}}.
$
Compute the denominator of that fraction:
$
1 - \frac{1}{x^2 + x + 1}
= \frac{x^2 + x + 1 - 1}{x^2 + x + 1}
= \frac{x^2 + x}{x^2 + x + 1}
= \frac{x(x + 1)}{x^2 + x + 1}.
$
Hence,
$
\tan(A + B)
= \frac{\frac{1 + x}{\sqrt{x} \,\sqrt{x^2 + x + 1}}}
{\frac{x(x + 1)}{x^2 + x + 1}}
= \frac{1 + x}{\sqrt{x} \,\sqrt{x^2 + x + 1}}
\times \frac{x^2 + x + 1}{x(x + 1)}.
$
The factor $ (1 + x) $ cancels with $ (x + 1) $, giving:
$
\tan(A + B)
= \frac{x^2 + x + 1}{\sqrt{x}\,\sqrt{x^2 + x + 1}\,x}
= \frac{\sqrt{x^2 + x + 1}}{x^{3/2}}.
$
Step 5: Compare with $ \tan C $
Rewrite $ \tan C $:
$
\tan C
= \sqrt{x^{-3} + x^{-2} + x^{-1}}
= \sqrt{\frac{1}{x^3} + \frac{1}{x^2} + \frac{1}{x}}
= \sqrt{\frac{x^2 + x + 1}{x^3}}
= \frac{\sqrt{x^2 + x + 1}}{\sqrt{x^3}}
= \frac{\sqrt{x^2 + x + 1}}{x^{3/2}}.
$
This matches $ \tan(A + B) $. Since $ \tan $ is one-to-one in $ (0, \tfrac{\pi}{2}) $, it implies:
$ A + B = C. $
Final Answer
The sum of the angles is equal to C.
