Question

lf $x\cos \alpha +y\sin \alpha =x\cos \beta +y\sin \beta =2a(0<\alpha ,\beta \pi /2)$, then

• A. $\cos \alpha +\cos \beta =\frac{4ax}{x^2+y^2}$
• B. $\cos \alpha \cos \beta =\frac{4a^2-y^2}{x^2+y^2}$
• C. $\sin \alpha +\sin \beta =\frac{4ay}{x^2+y^2}$
• D. $\sin \alpha \sin \beta =\frac{4a^2-x^2}{x^2+y^2}$

Answer 1

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

The correct options are
A $\cos \alpha +\cos \beta =\frac{4ax}{x^2+y^2}$
B $\cos \alpha \cos \beta =\frac{4a^2-y^2}{x^2+y^2}$
C $\sin \alpha +\sin \beta =\frac{4ay}{x^2+y^2}$
D $\sin \alpha \sin \beta =\frac{4a^2-x^2}{x^2+y^2}$

Multipyling first equation with $\cos \beta$ and second equation with $\cos \alpha$ we get

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

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

Similarly we get $x$ as

$x=\frac{2a\cos \left(\frac{\alpha +\beta }{2}\right)}{\cos \frac{\alpha -\beta }{2}}$

Hence,

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

Threfore ${\left(\cos \frac{\alpha -\beta }{2}\right)}^2=\frac{4a^2}{x^2+y^2}$

Multiplying both sides with x, we get

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

$\cos \alpha +\cos \beta =\frac{4ax}{x^2+y^2}$

Similarly

$\sin \alpha +\sin \beta =\frac{4ay}{x^2+y^2}$

Now

$\frac{4a^2-x^2}{x^2+y^2}$

$=\frac{4a^2}{x^2+y^2}-\frac{x^2}{x^2+y^2}$

$=\frac{4a^2}{x^2+y^2}-\frac{x^2}{x^2+y^2}$

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

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

$=\left(2\cos \alpha \cos \beta \right)(2\sin \alpha \cdot \sin \beta )$

$=\sin \alpha \cdot \sin \beta$

Similarly

$\frac{4a^2-y^2}{x^2+y^2}$

$=\cos \alpha \cdot \cos \beta$.