Question

If $tanA$ and $tanB$ are the roots of the equation $x^2-ax+b=0$, then the value of $si{n}^2(A+B)$ is

• A. $\frac{a^2}{a^2+(1-b{)}^2}$
• B. $\frac{a^2}{a^2+b^2}$
• C. $\frac{a^2}{(a^2+b^2)}$
• D. $\frac{a^2}{b^2+(1-a{)}^2}$

Answer 1

Answer: (A)

$\frac{a^2}{a^2+(1-b{)}^2}$

$\tan A$ and $\tan B$ are the roots of the equation $x^2-ax+b=0$

$\Rightarrow \tan A+\tan B=a,\tan A\tan B=b$

$\tan A+\tan B=a$

$\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}=a$

$\frac{\sin A\cos B+\sin B\cos A}{\cos A\cos B}=a$

$\sin (A+B)=a\cos A\cos B$

${\sin }^2(A+B)=a^2{\cos }^{2}a{\cos }^{2}B$

${\sin }^2(A+B)=\frac{a^2}{{\sec }^{2}A{\sec }^{2}B}$

${\sin }^2(A+B)=\frac{a^2}{(1+{\tan }^{2}A\left)\right(1+{\tan }^{2}B)}$

${\sin }^2(A+B)=\frac{a^2}{1+{\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}A{\tan }^{2}B}$

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

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

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

So option $A$ is correct.