If $tan A, tan B$ are the roots of $x^{2} - 2 x + 2 = 0$ then ${tan}^{2} (A + B) =$

4

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\frac{1}{2}

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\frac{3}{4}

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\frac{1}{4}

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Solution

The correct option is A $4$
tan A and tan B are roots of the given quadratic equation.

$\Rightarrow tan A + tan B = 2$ and $tan A \times tan B = 2$

We expand the expression to be evaluated as

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

substituting the values we get

$= {(\frac{2}{1 - 2})}^{2} = 4$

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