Question

lf $\alpha ,\beta$ are solutions of ${\sin }^{2}x+a\sin x+b=0$ and ${\cos }^{2}x+c\cos x+d=0$ then $\sin (\alpha +\beta )$ equals:-

• A. $\frac{2ac}{a^2+c^2}$
• B. $\frac{a^2+c^2}{2ac}$
• C. $\frac{2bd}{b^2+d^2}$
• D. $\frac{b^2+d^2}{2bd}$

Answer 1

The correct option is A $\frac{2ac}{a^2+c^2}$

Let $\alpha$ and $\beta$ are roots of the given equation ${\sin }^{2}x+a\sin x+b=0$. Then,

$\Rightarrow$ $\sin \alpha +\sin \beta =\frac{-a}{1}=-a$ .......... (A)

$\Rightarrow$ $\sin \alpha \sin \beta =\frac{b}{1}=b$

Let $\alpha$ and $\beta$ are roots of the given equation ${\cos }^{2}x+c\cos x+d=0$. Then,

$\Rightarrow$ $\cos \alpha +\cos \beta =\frac{-c}{1}=-c$ .......... (B)

$\Rightarrow$ $\cos \alpha \cos \beta =\frac{d}{1}=d$

$a^2+c^2=(\sin \alpha +\sin \beta {)}^2+(\cos \alpha +\cos \beta {)}^2$

$={\sin }^2\alpha +{\sin }^2\beta +2\sin \alpha \sin \beta +{\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta$

$=1+1+2(\sin \alpha \sin \beta +\cos \alpha \cos \beta )$

$=2[1+\cos (\alpha -\beta \left)\right]$

$=2\times 2{\cos }^2\left(\frac{\alpha -\beta }{2}\right)$

$=4{\cos }^2\left(\frac{\alpha -\beta }{2}\right)$

$\therefore$ $a^2+c^2$ $=4{\cos }^2\left(\frac{\alpha -\beta }{2}\right)$ ---- ( 1 )

Now, from A and B

$2\sin \frac{\alpha +\beta }{2}.\cos \frac{\alpha -\beta }{2}=-a$ --- ( 2 )

$2\cos \frac{\alpha +\beta }{2}.\cos \frac{\alpha -\beta }{2}=-c$ ---- ( 3 )

Multiplying ( 2 ) and ( 3 ),

$\Rightarrow$ $4\sin \frac{\alpha +\beta }{2}.\cos \frac{\alpha +\beta }{2}.{\cos }^2\left(\frac{\alpha -\beta }{2}\right)=ac$

$\Rightarrow$ $2\sin (\alpha +\beta )\left(\frac{a^2+c^2}{4}\right)=ac$ [ From ( 1 ) ]

$\Rightarrow$ $\sin (\alpha +\beta )=\frac{2ac}{a^2+c^2}$