Question

${\cos }^2\theta +{\cos }^2(\alpha +\theta )-2\cos \alpha \cos \theta {\cos }^2(\alpha +\theta )$ is independent of $\theta$.
True=7 False=9

Answer 1

${\cos }^2\theta +{\cos }^2(\alpha +\theta )-2\cos \alpha \cos \theta {\cos }^2(\alpha +\theta )$
$={\cos }^2\theta +\cos (\alpha +\theta )[\cos (\alpha +\theta )-2\cos \alpha \cos \theta ]$
$={\cos }^2\theta +\cos (\alpha +\theta )[\cos \alpha \cos \theta -\sin \alpha \sin \theta -2\cos \alpha \cos \theta ]$
$={\cos }^2\theta -\cos (\alpha +\theta )[\cos \alpha \cos \theta +\sin \alpha \sin \theta ]$
$={\cos }^2\theta -\cos (\alpha +\theta )\cos (\alpha -\theta )$
$={\cos }^2\theta -[{\cos }^2\alpha -{\sin }^2\theta ]={\cos }^2\theta +{\sin }^2\theta -{\cos }^2\alpha$
$=1-{\cos }^2\alpha$, which is independent of $\theta$
Ans: 7