Question

If $\tan \alpha$ and $\tan \beta$ are the roots of $x^2+ax+b=0$ then $\frac{\sin \left(\alpha +\beta \right)}{\sin \alpha sin\beta }$ is equal to

• A. $\frac{b}{a}$
• B. $\frac{a}{b}$
• C. $-\frac{a}{b}$
• D. $-\frac{b}{a}$

Answer 1

Answer: (C)

$-\frac{a}{b}$

$\tan \alpha$ and $\tan \beta$ are the roots of $x^2+ax+b=0$

$\Rightarrow$ $\tan \alpha +\tan \beta =\frac{-a}{1}$

$\Rightarrow$ $\frac{\sin \alpha }{\cos \alpha }+\frac{\sin \beta }{\cos \beta }=-a$

$\Rightarrow$ $\frac{\sin \alpha \cos \beta +\sin \beta \cos \alpha }{\cos \alpha .\cos \beta }=-a$

$\therefore$ $\frac{\sin (\alpha +\beta )}{\cos \alpha \cos \beta }=-a$ ----- ( 1 )

$\Rightarrow$ Now, $\tan \alpha \tan \beta =\frac{b}{1}$

$\Rightarrow$ $\frac{\sin \alpha }{\cos \alpha }.\frac{\sin \beta }{\cos \beta }=b$ ------ ( 2 )

Now, dividing ( 1 ) by ( 2 ) we get,

$\Rightarrow$ $\frac{\sin (\alpha +\beta )}{\sin \alpha \sin \beta }=\frac{-a}{b}$