Question

If $a=\cos \alpha +\cos \beta -\cos (\alpha +\beta ),$

$b=4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\cos \left(\frac{\alpha +\beta }{2}\right)$ then $a-b$=

• A. $0$
• B. $1$
• C. $-1$
• D. $-2$

Answer 1

The correct option is C $-1$

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

$=\cos \alpha +\cos \beta -(1+\cos (\alpha +\beta \left)\right)\{\because 2{\cos }^2\theta =1+\cos 2\theta \}$

$=a-1$

$\text{and}a-b=a-(a-1)$

$=1$

Option B is correct.