Question

If the equation ${\sin }^2(\theta -\alpha )\cos \alpha ={\cos }^2(\theta -\alpha )\sin \alpha =m\sin \alpha \cos \alpha$ has solution for all permissible value of $\theta ,\alpha \in R$ then

• A. $|m|>\frac{1}{\sqrt{2}}$
• B. $|m|\ge \frac{1}{\sqrt{2}}$
• C. $|m|>\sqrt{2}$
• D. $|m|\ge \sqrt{2}$

Answer 1

The correct option is B $|m|\ge \frac{1}{\sqrt{2}}$

${\sin }^2(\theta -\alpha )\cos \alpha ={\cos }^2(\theta -\alpha )\sin \alpha =m\sin \alpha \cos \alpha \Rightarrow m=\frac{{\sin }^2(\theta -\alpha )}{\sin \alpha }=\frac{{\cos }^2(\theta -\alpha )}{\cos \alpha }\Rightarrow m=\frac{{\sin }^2(\theta -\alpha )+{\cos }^2(\theta -\alpha )}{\sin \alpha +\cos \alpha }\{\because \frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\}m=\frac{1}{\sin \alpha +\cos \alpha }$
(for $m$ to be defined $\tan \alpha \ne -1$)
$\therefore |m|\ge \frac{1}{\sqrt{2}}$