If tan A and tan B are the roots of the quadratic equation x2-p x+q=0, then sin 2(A+B) is equal to

Question

If tan A and tan B are the roots of the quadratic equation x2-p x+q=0, then sin 2(A+B) is equal to (A) (p2/p2+q2) (B) (p2/(p+q)2) (C) 1-(p/(1-q)2) (D) (p2/p2+(1-q)2)

Tardigrade Admin · Accepted Answer

Since, tan A and tan B are the roots of the equation x2-p x+q=0 ∴ tan A+ tan B=p and tan A tan B=q ∴ tan (A+B) =( tan A+ tan B/1- tan A tan B) =(p/1-q) ⇒ sin (A+B) =(p/√p2+(1-q)2) ∴ sin 2(A+B)=(p2/p2+(1-q)2)

Q. If $tan A$ and $tan B$ are the roots of the quadratic equation $x^{2} - p x + q = 0$ , then $sin^{2} (A + B)$ is equal to

$\frac{p ^{2}}{p ^{2} + q ^{2}}$

$\frac{p ^{2}}{( p + q ) ^{2}}$

$1 - \frac{p}{( 1 - q ) ^{2}}$

$\frac{p ^{2}}{p ^{2} + ( 1 - q ) ^{2}}$

Q. If tanA and tanB are the roots of the quadratic equation x2−px+q=0, then sin2(A+B) is equal to

p2+q2p2​

(p+q)2p2​

1−(1−q)2p​

p2+(1−q)2p2​

Since, tanA and tanB are the roots of the equation x2−px+q=0 ∴tanA+tanB=p and tanAtanB=q ∴tan(A+B)=1−tanAtanBtanA+tanB​ =1−qp​ ⇒sin(A+B)=p2+(1−q)2​p​ ∴sin2(A+B)=p2+(1−q)2p2​

Q. If $tan A$ and $tan B$ are the roots of the quadratic equation $x^{2} - p x + q = 0$ , then $sin^{2} (A + B)$ is equal to

$\frac{p ^{2}}{p ^{2} + q ^{2}}$

$\frac{p ^{2}}{( p + q ) ^{2}}$

$1 - \frac{p}{( 1 - q ) ^{2}}$

$\frac{p ^{2}}{p ^{2} + ( 1 - q ) ^{2}}$