Question

If $\alpha$ and $\beta$ be the solution of $a\cos \theta +b\sin \theta =c$, then

• A. $\sin \alpha +\sin \beta =\frac{2bc}{a^2+b^2}$
• B. $\sin \alpha .\sin \beta =\frac{c^2-a^2}{a^2+b^2}$
• C. $\sin \alpha +\sin \beta =\frac{2ac}{c^2+b^2}$
• D. $\sin \alpha .\sin \beta =\frac{a^2-b^2}{b^2+c^2}$

Answer 1

Answer: (A), (B)

The correct options are
A $\sin \alpha +\sin \beta =\frac{2bc}{a^2+b^2}$
B $\sin \alpha .\sin \beta =\frac{c^2-a^2}{a^2+b^2}$

We have $a\cos \theta =c-b\sin \theta$

Squaring both sides

$\Rightarrow a^2(1-{\sin }^2\theta )=c^2-2bc\sin \theta +b^2{\sin }^2\theta$

$\Rightarrow (a^2+b^2){\sin }^2\theta -2bc\sin \theta +(c^2-a^2)=0$

Since $\alpha$ and $\beta$ are the values of $\theta$ as given,

$\therefore$ roots of the above equation are $\sin \alpha$ and $\sin \beta$.

$\therefore \sin \alpha +\sin \beta =$ Sum of roots $=\frac{2bc}{a^2+b^2}$

and $\sin \alpha .\sin \beta =$ Product of roots $=\frac{c^2-a^2}{a^2+b^2}$