Question

State True or False:

If $\alpha +\beta =\gamma$, then ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1+2\cos \alpha \cos \beta \cos \gamma$.

• A. True
• B. False

Answer 1

The correct option is A True

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

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

$L.H.S={\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma -1$

$={\cos }^2\alpha +{\cos }^2\beta -1(1-{\cos }^2\gamma )$

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

$={\cos }^2\alpha +{\cos }^2\beta -{\sin }^2(\alpha +\beta )$ [ Since, $\alpha +\beta =\gamma$ ]

$={\cos }^2\alpha +{\cos }^2\beta -[{\sin }^2\alpha {\cos }^2\beta +{\cos }^2\alpha {\sin }^2\beta +2\sin \alpha \cos \alpha \sin \beta \cos \beta$

$={\cos }^2\alpha (1-{\sin }^2\beta )+{\cos }^2\beta (1-{\sin }^2\alpha )-2\sin \alpha \cos \alpha \sin \beta \cos \beta$

$={\cos }^2\alpha {\cos }^2\beta +{\cos }^2\beta {\cos }^2\alpha -2\sin \alpha \cos \alpha \sin \beta \cos \beta$

$=2{\cos }^2\alpha {\cos }^2\beta -2\sin \alpha \cos \alpha \sin \beta \cos \beta$

$=2\cos \alpha \cos \beta [\cos \alpha \cos \beta -\sin \alpha \sin \beta ]$

$=2\cos \alpha \cos \beta \cos (\alpha +\beta )$

$=2\cos \alpha \cos \beta \cos \gamma$

$=R.H.S$

Hence, we have proved that, ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma -1=2\cos \alpha \cos \beta \cos \gamma$