Question

$(\cos \alpha -\cos \beta {)}^2+(\sin \alpha -\sin \beta {)}^2=$ $k{\sin }^2\left(\frac{\alpha -\beta }{2}\right)$, then $k$ is equal to:

• A. $4$
• B. $3$
• C. $\frac{3}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $4$

LHS$=({\cos }^2\alpha +{\cos }^2\beta -2\cos \alpha \cos \beta )+({\sin }^2\alpha +{\sin }^2\beta -2\sin \alpha \sin \beta )$
$=2-2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )$
$=2-2\cos (\alpha -\beta )$
$=4{\sin }^2\frac{\alpha -\beta }{2}$