If tan A, tan B are the roots of $x^{2} - P x + Q = 0$ the value of $s i n^{2}$ (A+B)=(where P, Q $ϵ$ R)

$\frac{P^{2}}{P^{2} + Q^{2}}$

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$\frac{P^{2}}{(P + Q)^{2}}$

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$\frac{P^{2}}{P^{2} + (1 - Q)^{2}}$

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$\frac{Q^{2}}{P^{2} + (1 - Q)^{2}}$

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Solution

The correct option is C
$\frac{P^{2}}{P^{2} + (1 - Q)^{2}}$

$t a n (A + B) = \frac{P}{1 - Q} t h e n s i n^{2} (A + B) = \frac{t a n^{2} (A + B)}{1 + t a n^{2} (A + B)}$

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Similar questions

If tan A, tan B are the roots of $x^{2} - P x + Q = 0$ the value of $s i n^{2}$ (A+B)=(where P, Q $ϵ$ R)

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°

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