Question

If $x\cos \alpha +y\sin \alpha =x\cos \beta +y\sin \beta =a,2n\pi <\alpha ,\beta <(4n+1)\frac{\pi }{2},$ then

• A. $\cos \alpha +\cos \beta =\frac{2ax}{x^2+y^2}$
• B. $\sin \alpha +\sin \beta =\frac{2ay}{x^2+y^2}$
• C. $\cos \alpha \cos \beta =\frac{a^2-y^2}{x^2+y^2}$
• D. $\sin \alpha \sin \beta =\frac{x^2-a^2}{x^2+y^2}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\sin \alpha +\sin \beta =\frac{2ay}{x^2+y^2}$
B $\cos \alpha +\cos \beta =\frac{2ax}{x^2+y^2}$
C $\cos \alpha \cos \beta =\frac{a^2-y^2}{x^2+y^2}$

Evidently $\alpha and\beta$ are the roots of the equation $x\cos \theta +y\sin \theta =a$.....(1)

or$x^2{\cos }^2\theta =(a-y\sin \theta {)}^{2}or{x}^2-x^2{\sin }^2\theta =a^2+y^2{\sin }^2\theta -2ay\sin \theta$

or ${\sin }^2\theta (x^2+y^2)-2ay\sin \theta +a^2-x^2=0$

The roots of the equation are $\sin \alpha$ and $\sin \beta$

$\therefore \sin \alpha +\sin \beta =\frac{2ay}{x^2+y^2}$ and $\sin \alpha \sin \beta .=\frac{a^2-x^2}{x^2+y^2}$

Writing (1) in terms of $\cos \theta ,(x^2+y^2){\cos }^2\theta -2ax\cos \theta +a^2-y^2=0$

$\therefore \cos \alpha +\cos \beta =\frac{2ax}{x^2+y^2}and\cos \alpha \cos \beta =\frac{a^2-y^2}{x^2+y^2}$