Question

If $\alpha ,\beta$ 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 $\frac{2ac}{a^2+c^2}$

Given:

$\Rightarrow$$\alpha ,\beta$ are roots of the equation ${\sin }^{2}x+a\sin x+b=0$

$\Rightarrow$${\sin }^2\alpha +a\sin \alpha +b=0$.....$\left(i\right)$

$\Rightarrow$${\sin }^2\beta +a\sin \beta +b=0$........$\left(ii\right)$

Equation (i) - equation (ii)

$\Rightarrow$${\sin }^2\alpha -{\sin }^2\beta +a(\sin \alpha -\sin \beta )=0$

$\Rightarrow$$(\sin \alpha -\sin \beta )(\sin \alpha +\sin \beta +a)=0$

$\Rightarrow \sin \alpha +\sin \beta =-a$

$\Rightarrow$$2\sin \frac{\alpha +\beta }{2}\cdot \cos \frac{\alpha -\beta }{2}=-a$ ....(1)

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

$\Rightarrow$${\cos }^2\alpha +c\cos \alpha +b=0$$.....\left(iii\right)$

$\Rightarrow$${\cos }^2\beta =c\cos \beta +b=0$$.......\left(iv\right)$

Equation (iii) - Equation (iv)

$\Rightarrow$${\cos }^2\alpha -{\cos }^2\beta +b(\cos \alpha -\cos \beta )=0$

$\Rightarrow$$(\cos \alpha -\cos \beta )(\cos \alpha +\cos \beta +b)=0$

$\Rightarrow \cos \alpha +\cos \beta =-c$

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

On dividing equation(1) by (2), we get

$\therefore \tan \frac{\alpha +\beta }{2}=\frac{a}{c}$
$\therefore \sin (\alpha +\beta )=\frac{2\tan \left(\frac{\alpha +\beta }{2}\right)}{1+{\tan }^2\frac{\alpha +\beta }{2}}$
$=\frac{2ac}{a^2+c^2}$