Given that tan A, tan B are the roots of the equation $x^{2}$ - px + q = 0, then the value of $s i n^{2}$ (A + B) is

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Solution

The correct option is A

tan A + tan B = p, tan A tan B = q

tan(A + B) = $\frac{p}{1 - q}$ , $s i n^{2}$ (A + B) = $t a n^{2}$ (A + B) $s e c^{2}$ (A + B)

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Similar questions

Q. Given that

t a n A, t a n B

are the roots of the equation

x^{2} - p x + q = 0

, then the value of

s i n^{2} (A + B)

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)

$(q + p) \times (q - p) = ?$

Q. Assertion :If tan A, tan B are the roots of equation

x^{2} - p x - 1 = 0

, then

{sin}^{2} (A + B) = \frac{p^{2}}{1 + p^{2}}

Reason:

{sin}^{2} (A + B) = \frac{t a n^{2} (A + B)}{1 + t a n^{2} (A + B)}

If $α, β$ are the roots of the equation $x^{2} + p x + 1 = 0; γ, δ$ the roots of the equation $x^{2} + p x + 1 = 0;$ then $(α - γ) (α + δ) (β - γ) (β + δ)$

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Given that tan A, tan B are the roots of the equation x2 - px + q = 0, then the value of sin2(A + B) is

The correct option is A tan A + tan B = p, tan A tan B = q tan(A + B) = p1−q, sin2(A + B) = tan2(A + B) sec2(A + B)

Given that tan A, tan B are the roots of the equation $x^{2}$ - px + q = 0, then the value of $s i n^{2}$ (A + B) is

The correct option is A

tan A + tan B = p, tan A tan B = q

tan(A + B) = $\frac{p}{1 - q}$ , $s i n^{2}$ (A + B) = $t a n^{2}$ (A + B) $s e c^{2}$ (A + B)