If $tan A$ and $tan B$ are the roots of quadratic equation $x^{2} - a x + b = 0,$ then the value of ${sin}^{2} (A + B)$ is

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Solution

$x^{2} - a x + b = 0$ has roots $tan A, tan B$
So, $tan A + tan B = a$
and $tan A \cdot tan B = b$
$tan (A + B) = \frac{tan A + tan B}{1 - tan A \cdot tan B} \Rightarrow tan (A + B) = \frac{a}{1 - b}$

Therefore, ${sin}^{2} (A + B)$
$= \frac{1}{{cosec}^{2} (A + B)} = \frac{1}{1 + {cot}^{2} (A + B)} = \frac{1}{1 + {(\frac{1 - b}{a})}^{2}} = \frac{a^{2}}{a^{2} + (1 - b^{2})}$

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