Question

The points $O(0,0),A(\cos \alpha ,\sin \alpha )$ and $B(\cos \beta ,\sin \beta )$ are the vertices of a right-angled triangle if

• A. $\sin \frac{\alpha -\beta }{2}=\frac{1}{\sqrt{2}}$
• B. $\cos \frac{\alpha -\beta }{2}=-\frac{1}{\sqrt{2}}$
• C. $\cos \frac{\alpha -\beta }{2}=\frac{1}{\sqrt{2}}$
• D. $\sin \frac{\alpha -\beta }{2}=-\frac{1}{\sqrt{2}}$

Answer 1

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

The correct options are
A $\cos \frac{\alpha -\beta }{2}=-\frac{1}{\sqrt{2}}$
B $\sin \frac{\alpha -\beta }{2}=\frac{1}{\sqrt{2}}$
C $\cos \frac{\alpha -\beta }{2}=\frac{1}{\sqrt{2}}$
D $\sin \frac{\alpha -\beta }{2}=-\frac{1}{\sqrt{2}}$

$OA=1$ and $OB=1$ $\therefore AB=\sqrt{2}$
$\therefore (\cos \beta -\cos \alpha {)}^2+(\sin \beta -\sin \alpha {)}^2=2$
$⟹{\cos }^2\alpha +{\sin }^2\alpha +{\cos }^2\beta +{\sin }^2\beta -2\sin \beta \sin \alpha -2\cos \beta \cos \alpha =2$
$⟹2\sin \beta \sin \alpha +2\cos \beta \cos \alpha =0$
$\therefore \cos (\alpha -\beta )=0$
$⟹\sin \left(\frac{\alpha -\beta }{2}\right)=\pm \frac{1}{\sqrt{2}}$
$⟹\cos \left(\frac{\alpha -\beta }{2}\right)=\pm \frac{1}{\sqrt{2}}$