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$\tan 22^{0}$ and $\tan 23^{0}$ are roots $x^{2} + a x + b = 0$ , then

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$a + b = 1$

$a - b + 1 = 0$

$a + b + 1 = 0$

$b - a + 1 = 0$

answer is B.

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Detailed Solution

Let $α = \tan 22, β = \tan 23$
$x^{2} + a x + b = 0 \Rightarrow α + β = - a$ , $α \times β = b$
$\tan 45 = \frac{\tan 22 + \tan 23}{1 - \tan 22 \tan 23} = \frac{α + β}{1 - α β}$

$1 = \frac{- a}{1 - b} \Rightarrow 1 - b = - a \Rightarrow a - b + 1 = 0$

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