Question

If $\alpha +\beta =\gamma$ then $co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma -2cos\alpha cos\beta cos\gamma$ =

• A. 1
• B. 0
• C. -1
• D. 2

Answer 1

The correct option is A 1

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma -2\cos \alpha \cos \beta \cos \gamma$

$\frac{1+\cos 2\alpha +1+\cos 2\beta }{2}+{\cos }^2\gamma -2\cos \alpha \cos \beta \cos \gamma$

$\frac{1+\cos 2\alpha +1+\cos 2\beta +2{\cos }^2\gamma }{2}-2\cos \alpha \cos \beta \cos \gamma$

$\frac{2+\cos 2\alpha +\cos 2\beta +2{\cos }^2\gamma }{2}-2\cos \alpha \cos \beta \cos \gamma$

$\frac{2+2\cos (\alpha +\beta )\cos (\alpha -\beta )+2{\cos }^2\gamma }{2}-2\cos \alpha \cos \beta \cos \gamma$

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

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

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

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

$\frac{2+4\cos \alpha \cos \beta \cos \gamma }{2}-2\cos \alpha \cos \beta \cos \gamma$

$1+2\cos \alpha \cos \beta \cos \gamma -2\cos \alpha \cos \beta \cos \gamma$

$1$