Question

If $\alpha ,\beta$ are solution of equation a $cos\theta +bsin\theta =c$ then

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

Answer 1

The correct option is D $\sin \alpha .\sin \beta =\frac{a^2-c^2}{b^2-a^2}$

$a\cos \theta +b\sin \theta =c$

$a^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab{\sin }^2\theta .\cos \theta =c^2$

$a^2(1-{\sin }^2\theta )+b^2{\sin }^2\theta +2ab\sin \theta .\cos \theta =c^2$

$a^2-a^2{\sin }^2\theta +b^2{\sin }^2\theta {\sin }^2\theta +2ab\sin \theta \cos \theta =c^2$

${\sin }^2\theta (b^2-a^2)+2ab\sin \theta .\cos \theta +a^2-c^2=0$

So

$\sin \alpha .\sin \beta =\frac{a^2-c^2}{b^2-a^2}$